Chapter 10 · Thermal Properties of Matter

10.2 Specific Heat, Calorimetry, Change of State and Latent Heat

How much heat does it take to warm a body, melt it or boil it? Specific heat capacity answers the first ($Q=mc\,\Delta T$), latent heat the other two ($Q=mL$), and the method of mixtures measures both.

Put a steel spoon and a cup of water on the same stove flame for one minute. The spoon becomes too hot to touch; the water is barely warm. On a summer afternoon the sand on a beach burns your feet while the sea stays cool, and at night the sea is the warmer of the two. Different substances need very different amounts of energy to change their temperature by the same amount — and water needs more than almost anything else, which is why the oceans moderate the climate of every coastal city from Mumbai to Lisbon.

This section does the bookkeeping of heat. We define the heat capacity of a body and the specific heat capacity of a substance, and use energy conservation in the method of mixtures to measure them in a calorimeter. Then we look at what happens when heating no longer raises the temperature at all: ice melting at $0\ ^\circ\text{C}$, water boiling at $100\ ^\circ\text{C}$. The energy that goes into these changes of state is the latent heat, and it is large — boiling a kettle dry takes more than five times the energy needed to bring it to the boil.

Heat capacity and specific heat capacity

Supply heat $\Delta Q$ to a body, with no change of state, and its temperature rises by $\Delta T$. Doubling the heat doubles the rise; doubling the mass of the body halves it. The proportionality constant depends on what the body is made of.

📘 Definition 10.5 — Heat capacity and specific heat capacity

The heat capacity $S$ of a body is the heat needed per unit rise in its temperature, and the specific heat capacity $c$ of a substance is the heat capacity per unit mass:

$$S=\frac{\Delta Q}{\Delta T}\ \ (\text{J/K}),$$ $$c=\frac{1}{m}\frac{\Delta Q}{\Delta T}\ \ (\text{J kg}^{-1}\text{K}^{-1}).$$

Heat capacity belongs to a particular object (this kettle, that brick); specific heat capacity belongs to the material.

📐 Result 10.7 — Heat for a temperature change

To change the temperature of a mass $m$ of a substance by $\Delta T$, without a change of state and over a range in which $c$ is nearly constant, the heat required is

$$Q=mc\,\Delta T=S\,\Delta T .$$

$Q\gt0$ (heat absorbed) for a rise in temperature and $Q\lt0$ (heat given out) for a fall.

Substance$c$ ($\text{J kg}^{-1}\text{K}^{-1}$)Substance$c$ ($\text{J kg}^{-1}\text{K}^{-1}$)
water (liquid)4186aluminium900
ice (near $0\ ^\circ\text{C}$)2100iron / steel450
steam (near $100\ ^\circ\text{C}$)2010copper387
mercury140lead128

Water's value, $4186\ \text{J kg}^{-1}\text{K}^{-1}$, is about ten times that of iron and thirty times that of lead. That is why water is used in car radiators and hot-water bottles, why a sea breeze blows onshore by day (the land heats faster than the sea), and why islands and coasts have mild climates. Notice also that ice and steam have only about half the specific heat capacity of liquid water: the same substance in different states has different values.

📘 Definition 10.6 — Molar specific heat capacity

The molar specific heat capacity of a substance is the heat capacity per mole:

$$C=\frac{1}{n}\frac{\Delta Q}{\Delta T}\ \ (\text{J mol}^{-1}\text{K}^{-1}),\qquad C=Mc ,$$

where $M$ is the molar mass in kg/mol. For a solid or liquid the value hardly depends on how it is heated; for a gas it depends strongly on whether the pressure or the volume is held fixed, giving two values $C_p$ and $C_v$ (Chapter 11).

✏️ Example 1 — Boiling a kettle
An electric kettle rated at $2.0\ \text{kW}$ holds $1.5\ \text{kg}$ of water at $20\ ^\circ\text{C}$. (a) How much heat is needed to bring the water to $100\ ^\circ\text{C}$? (b) Ignoring heat losses and the kettle itself, how long does it take? (c) If the same heat went into $1.5\ \text{kg}$ of aluminium instead, by how much would its temperature rise?
(a) $Q=mc\Delta T=1.5\times4186\times80=5.02\times10^5\ \text{J}$.
(b) Power is energy per unit time (Section 5.3): $t=\dfrac{Q}{P}=\dfrac{5.02\times10^5}{2.0\times10^3}=251\ \text{s}\approx4.2\ \text{min}$.
(c) $\Delta T=\dfrac{Q}{mc}=\dfrac{5.02\times10^5}{1.5\times900}=372\ \text{K}$ — starting from $20\ ^\circ\text{C}$, the aluminium would reach about $390\ ^\circ\text{C}$.
Check The ratio of the rises, $\dfrac{372}{80}=4.65$, equals the ratio of the specific heats, $\dfrac{4186}{900}=4.65$ ✓. A real kettle takes a little longer: some heat warms the kettle body and leaks to the air.
✏️ Example 2 — Molar heat capacity of copper
(a) Find the molar specific heat capacity of copper, molar mass $63.5\ \text{g/mol}$, and compare it with $3R$. (b) Find the heat capacity of a $2.0\ \text{kg}$ copper block, and the heat it gives out when it cools by $15\ \text{K}$.
(a) $C=Mc=0.0635\times387=24.6\ \text{J mol}^{-1}\text{K}^{-1}$, while $3R=3\times8.31=24.9\ \text{J mol}^{-1}\text{K}^{-1}$ — within about $1\%$.
(b) $S=mc=2.0\times387=774\ \text{J/K}$. Cooling by $15\ \text{K}$: $Q=S\Delta T=774\times(-15)=-1.16\times10^4\ \text{J}$, i.e. it gives out $1.2\times10^4\ \text{J}$.
Note Per kilogram, copper and aluminium differ by a factor of more than two; per mole, most metals at room temperature have $C\approx3R\approx25\ \text{J mol}^{-1}\text{K}^{-1}$. Chapter 12 explains this from the energy of vibrating atoms.

Calorimetry: the method of mixtures

Drop a hot lump of metal into cold water and stir. The metal cools, the water warms, and soon both settle at one common temperature. If no energy escapes to the surroundings, every joule the metal gives out is taken in by the water. Measuring temperatures, we can deduce heats — this is calorimetry, "heat measuring".

📐 Principle 10.8 — Principle of calorimetry

When bodies at different temperatures are mixed in a thermally isolated system (no heat exchanged with the surroundings and no work done), they come to a common final temperature and

$$Q_\text{lost}=Q_\text{gained},$$

the heat lost by the hotter bodies equalling the heat gained by the colder ones;

or equivalently $\sum Q_i=0$ with each $Q_i=m_ic_i\left(T_f-T_i\right)$ taken with its sign. It is the conservation of energy applied to heat.

A calorimeter makes the isolation as good as possible: a thin copper vessel with a copper stirrer, sitting on insulating supports inside an outer jacket, with a lid through which a thermometer passes. Copper is chosen because it conducts well (the vessel quickly reaches the temperature of its contents) and has a small specific heat; polishing it reduces radiation. The vessel itself absorbs heat, so it is included in the balance. Its water equivalent is the mass of water with the same heat capacity: $W=\dfrac{m_\text{cal}c_\text{cal}}{c_\text{water}}$.

🧭 Problem-solving strategy — method of mixtures
  1. List every body: mass, specific heat capacity, initial temperature. Don't forget the calorimeter and stirrer.
  2. Decide (or guess) whether any change of state happens — see the latent-heat strategy later in this section.
  3. Write heat lost $=$ heat gained, each as $mc\times$(a positive temperature difference) — or write $\sum mc(T_f-T_i)=0$ and let the signs look after themselves.
  4. Solve for the unknown; check that $T_f$ lies between the initial temperatures and that the heats balance numerically.
✏️ Example 3 — Predicting the final temperature
A $0.200\ \text{kg}$ block of aluminium is taken from boiling water at $100\ ^\circ\text{C}$ and dropped into $0.150\ \text{kg}$ of water at $20\ ^\circ\text{C}$ in a copper calorimeter of mass $0.100\ \text{kg}$, also at $20\ ^\circ\text{C}$. Find the final temperature, and check the heat balance.
Heat capacities Aluminium: $0.200\times900=180\ \text{J/K}$. Water $+$ calorimeter: $0.150\times4186+0.100\times387=627.9+38.7=666.6\ \text{J/K}$.
Balance $180(100-T_f)=666.6(T_f-20)$, so $T_f=\dfrac{180\times100+666.6\times20}{180+666.6}=\dfrac{31\,332}{846.6}=37.0\ ^\circ\text{C}$.
Check Lost by aluminium: $180\times63.0=1.13\times10^4\ \text{J}$. Gained: water $627.9\times17.0=1.07\times10^4\ \text{J}$, calorimeter $38.7\times17.0=658\ \text{J}$; total $1.13\times10^4\ \text{J}$ ✓.
Note The final temperature is a heat-capacity-weighted average of the starting temperatures — closer to $20\ ^\circ\text{C}$ because the water's heat capacity is nearly four times the block's. This is the default of Interactive 1.
✏️ Example 4 — Measuring a specific heat capacity
A $0.200\ \text{kg}$ piece of an unknown metal at $100.0\ ^\circ\text{C}$ is placed in $0.150\ \text{kg}$ of water at $20.0\ ^\circ\text{C}$, in a $0.100\ \text{kg}$ copper calorimeter. The final temperature is $30.0\ ^\circ\text{C}$. Find the specific heat capacity of the metal. Which metal in the table might it be?
Gained Water and calorimeter, $666.6\ \text{J/K}$, warm by $10.0\ \text{K}$: $6666\ \text{J}$.
Lost The metal cools by $70.0\ \text{K}$: $0.200\times c\times70.0=14.0\,c$.
Solve $14.0\,c=6666$, so $c=476\ \text{J kg}^{-1}\text{K}^{-1}\approx4.8\times10^2\ \text{J kg}^{-1}\text{K}^{-1}$ — close to iron or steel ($450$).
Errors Heat that leaks to the air while the metal is transferred, or while the mixture settles, makes $T_f$ lower than it should be, so the deduced $c$ comes out too small. A lid, good stirring and a quick transfer reduce this. (Steels and cast irons genuinely range from about $420$ to $500\ \text{J kg}^{-1}\text{K}^{-1}$, so $476$ is a plausible steel.)
Interactive 1 — Method-of-Mixtures Calorimeter

A hot metal block (red) is dropped into water at $20\ ^\circ\text{C}$ in a $0.100\ \text{kg}$ copper calorimeter (blue: water $+$ calorimeter). Press Play: heat flows from hot to cold, fastest at first, until the two temperatures meet at $T_f$. The bars on the right show the heat lost by the metal and the heat gained by the water and calorimeter — always equal. Change the metal, the masses and the starting temperature.

200 g100 °C150 g
final temperature $T_f$—
heat lost by metal—
gained by water—
gained by calorimeter—

Change of state

Matter is usually met as a solid, a liquid or a gas. Heat a block of ice from $-10\ ^\circ\text{C}$ on a steady flame, with a thermometer in it, and record the temperature every half-minute. It rises steadily to $0\ ^\circ\text{C}$ — and then stops. For minutes the thermometer reads $0\ ^\circ\text{C}$ while the ice melts, although the flame is still supplying heat. Only when the last crystal has melted does the temperature rise again, now more slowly (liquid water has the larger specific heat), until at $100\ ^\circ\text{C}$ it stops again, this time for much longer, while the water boils away.

📘 Definition 10.7 — Changes of state; melting and boiling points

A change of state is a transition between solid, liquid and gas: melting (fusion) and its reverse freezing; vaporisation (boiling) and its reverse condensation; and sublimation, solid straight to vapour (camphor, iodine, solid carbon dioxide).

The melting point is the temperature at which the solid and liquid states coexist in equilibrium; the boiling point is the temperature at which the liquid and its vapour coexist with the vapour pressure equal to the pressure above the liquid. At standard atmospheric pressure these are the normal melting and boiling points. While a pure substance changes state, its temperature stays constant.

Both points depend on pressure. Raising the pressure on water raises its boiling point: in a pressure cooker at about twice atmospheric pressure, water boils at about $120\ ^\circ\text{C}$ and food cooks much faster. Lowering the pressure lowers it: at Leh, about $3500\ \text{m}$ up, water boils near $88\ ^\circ\text{C}$, and lentils take far longer to cook. For ice the effect of pressure is reversed — because ice is less dense than water, squeezing it lowers its melting point (by about $0.007\ \text{K}$ per atmosphere). A thin wire carrying heavy weights, slung over a block of ice, sinks slowly through it: the ice under the wire melts at the high pressure, the water flows over the wire and refreezes above it. This melting under pressure and refreezing is called regelation. Only at the triple point (Section 10.1) can solid, liquid and vapour all coexist.

Boiling and evaporation. Evaporation happens at the surface of a liquid at any temperature, as the fastest molecules escape; boiling happens throughout the liquid, at one temperature, as bubbles of vapour form inside it. Evaporation cools what is left behind — which is how sweat cools you and an earthen pot (matka) keeps water cool.

Latent heat

During a change of state the heat supplied does not raise the temperature: it is used to pull the molecules apart against their mutual attraction (and, on boiling, to push back the atmosphere as the vapour expands). It is "hidden" from the thermometer — hence the name latent heat, from the Latin for "hidden".

📘 Definition 10.8 — Latent heat

The latent heat $L$ of a change of state is the heat required per unit mass to change the state of a substance at constant temperature and pressure:

$$L=\frac{Q}{m}\ \ (\text{J/kg}).$$

For solid → liquid it is the latent heat of fusion $L_f$; for liquid → vapour, the latent heat of vaporisation $L_v$. The same heat is given out in the reverse change (freezing, condensation). For water at standard pressure, $L_f=3.33\times10^5\ \text{J/kg}$ and $L_v=2.26\times10^6\ \text{J/kg}$.

📐 Result 10.9 — Heat for a change of state

The heat absorbed when a mass $m$ melts (or boils) completely at its melting (or boiling) point is

$$Q=mL ,$$

with the temperature remaining constant throughout. The heat given out on freezing (or condensing) is the same, $mL$.

For water $L_v$ is almost seven times $L_f$: melting only loosens the molecules, which stay in contact; boiling separates them completely. This is why a scald from steam is far worse than one from boiling water — each gram of steam that condenses on the skin releases $2.26\ \text{kJ}$ before it even starts to cool.

✏️ Example 5 — From ice to steam
How much heat is needed to turn $0.200\ \text{kg}$ of ice at $-10\ ^\circ\text{C}$ into steam at $100\ ^\circ\text{C}$? What fraction of it goes into boiling? Take $c_\text{ice}=2100\ \text{J kg}^{-1}\text{K}^{-1}$.
Warm the ice $Q_1=mc_\text{ice}\Delta T=0.200\times2100\times10=4.20\times10^3\ \text{J}$.
Melt it $Q_2=mL_f=0.200\times3.33\times10^5=6.66\times10^4\ \text{J}$.
Heat the water $Q_3=mc_\text{w}\Delta T=0.200\times4186\times100=8.37\times10^4\ \text{J}$.
Boil it $Q_4=mL_v=0.200\times2.26\times10^6=4.52\times10^5\ \text{J}$.
Total $Q=4200+66\,600+83\,720+452\,000=6.07\times10^5\ \text{J}$; boiling takes $\dfrac{452\,000}{606\,520}=74.5\%$ of it.
Check Interactive 2 starts from these values: drag the energy to $607\ \text{kJ}$ and the last drop has just boiled. The two flat stretches of its graph are $66.6$ and $452\ \text{kJ}$ long.
Interactive 2 — Heating Curve: Ice to Steam

A sample of ice is heated steadily. Drag the energy supplied (or press Play for a constant-power heater). The graph plots temperature against energy: sloping parts are $Q=mc\,\Delta T$ — steeper where $c$ is smaller — and the flat plateaus are the latent heats, $Q=mL$, during which the temperature does not change. The beaker shows how much is ice, water and steam.

—
energy supplied—
temperature—
state—

Mixtures with a change of state. When ice is put into warm water, you do not know in advance whether all the ice will melt. Guessing wrong gives nonsense — a "final temperature" below $0\ ^\circ\text{C}$ for a mixture that contains liquid water, or above $0\ ^\circ\text{C}$ with ice still floating in it. The safe method is to compare heats first.

🧭 Problem-solving strategy — ice (or steam) in a mixture
  1. Find $Q_\text{avail}$: the heat the warm bodies would give out in cooling to $0\ ^\circ\text{C}$.
  2. Find $Q_\text{need}$: the heat needed to bring all the ice to $0\ ^\circ\text{C}$ and melt it completely.
  3. If $Q_\text{avail}\gt Q_\text{need}$: all the ice melts; now solve the ordinary mixture problem with the melted ice as water at $0\ ^\circ\text{C}$.
  4. If $Q_\text{avail}\lt Q_\text{need}$: the final temperature is $0\ ^\circ\text{C}$ and only part of the ice melts: $m_\text{melted}=\dfrac{Q_\text{avail}-Q_\text{warm ice to }0}{L_f}$.
  5. Steam is handled the same way with $100\ ^\circ\text{C}$ and $L_v$.
✏️ Example 6 — Not all the ice melts
$0.100\ \text{kg}$ of ice at $0\ ^\circ\text{C}$ is added to $0.300\ \text{kg}$ of water at $20\ ^\circ\text{C}$ in a vacuum flask whose own heat capacity is negligible. Find the final temperature and the mass of ice left.
Available Water cooling to $0\ ^\circ\text{C}$: $Q_\text{avail}=0.300\times4186\times20=2.51\times10^4\ \text{J}$.
Needed To melt all the ice: $Q_\text{need}=0.100\times3.33\times10^5=3.33\times10^4\ \text{J}$. Since $Q_\text{avail}\lt Q_\text{need}$, not all of it melts and the final temperature is $0\ ^\circ\text{C}$.
Melted $m=\dfrac{25\,116}{3.33\times10^5}=0.0754\ \text{kg}$; ice left $=0.100-0.0754=0.025\ \text{kg}$ ($25\ \text{g}$).
Wrong route Blindly writing $0.100L_f+0.100c_w(T-0)=0.300c_w(20-T)$ gives $T=-4.9\ ^\circ\text{C}$ — liquid water below its freezing point in a mixture with ice, which is impossible. The negative answer is the signal that the assumption "all ice melts" was false.
⚠️ Common mistakes

1. Heat capacity versus specific heat capacity. $S$ (J/K) is for an object; $c$ (J kg⁻¹ K⁻¹) is for a material. $S=mc$.

2. Using $mc\Delta T$ across a change of state. Split the process at $0\ ^\circ\text{C}$ and $100\ ^\circ\text{C}$: different $c$ for ice, water and steam, and an $mL$ term at each transition.

3. Forgetting the calorimeter. The vessel and stirrer absorb heat too.

4. Assuming all the ice melts. Compare $Q_\text{avail}$ with $Q_\text{need}$ first (Example 6).

5. "Heat makes the temperature rise." Not during melting or boiling — the heat goes into the change of state.

6. Grams and kilograms. With $c$ in $\text{J kg}^{-1}\text{K}^{-1}$ and $L$ in J/kg, masses must be in kg. In calorie units ($c_w=1\ \text{cal g}^{-1}\,{}^\circ\text{C}^{-1}$, $L_f=80\ \text{cal/g}$, $L_v=540\ \text{cal/g}$) use grams — never mix the two systems.

Exercises

Unless told otherwise use $c_\text{water}=4186$, $c_\text{ice}=2100$, $c_\text{steam}=2010\ \text{J kg}^{-1}\text{K}^{-1}$, $L_f=3.33\times10^5\ \text{J/kg}$, $L_v=2.26\times10^6\ \text{J/kg}$, the table of this section, and neglect heat exchange with the surroundings.

I
Specific heat capacity.
  • 1 How much heat is needed to warm $2.5\ \text{kg}$ of water from $15\ ^\circ\text{C}$ to $65\ ^\circ\text{C}$?
  • 2 A $500\ \text{g}$ aluminium frying pan is heated from $20\ ^\circ\text{C}$ to $180\ ^\circ\text{C}$. How much heat does it absorb?
  • 3 $3.0\ \text{kJ}$ of heat raises the temperature of a $0.40\ \text{kg}$ metal block by $20\ \text{K}$. Find its specific heat capacity.
  • 4 Find the heat capacity of a $0.25\ \text{kg}$ copper block, and its temperature rise when it absorbs $1.0\ \text{kJ}$.
  • 5 Find the molar specific heat capacity of aluminium (molar mass $27\ \text{g/mol}$) and compare it with $3R$.
  • 6 A $1.0\ \text{kW}$ immersion heater is placed in $2.0\ \text{kg}$ of water at $20\ ^\circ\text{C}$. (a) How long does it take to reach $80\ ^\circ\text{C}$? (b) How long if $15\%$ of the heater's output is lost to the surroundings?
  • 7 Explain why a city on the coast has smaller day–night and summer–winter temperature swings than a city far inland at the same latitude.
  • 8 (Modelling) A $10\ \text{g}$ lead bullet moving at $300\ \text{m/s}$ stops in a wooden block. If half of its kinetic energy heats the bullet, find its rise in temperature.
1 $Q=2.5\times4186\times50=5.2\times10^5\ \text{J}$ ($5.23\times10^5$).
2 $Q=0.500\times900\times160=7.2\times10^4\ \text{J}$.
3 $c=\dfrac{Q}{m\Delta T}=\dfrac{3000}{0.40\times20}=3.8\times10^2\ \text{J kg}^{-1}\text{K}^{-1}$ ($375$; probably copper or brass).
4 $S=0.25\times387=97\ \text{J/K}$; $\ \Delta T=\dfrac{1000}{96.75}=10\ \text{K}$ ($10.3$).
5 $C=Mc=0.027\times900=24.3\ \text{J mol}^{-1}\text{K}^{-1}$, close to $3R=24.9$.
6 (a) $Q=2.0\times4186\times60=5.02\times10^5\ \text{J}$; $t=\dfrac{5.02\times10^5}{1000}=502\ \text{s}\approx8.4\ \text{min}$. (b) Only $850\ \text{W}$ heats the water: $t=\dfrac{5.02\times10^5}{850}=591\ \text{s}\approx9.8\ \text{min}$.
7 The sea's huge mass and high specific heat capacity mean it changes temperature very little as it absorbs heat by day (and in summer) and gives it back at night (and in winter). Air over the coast is kept close to the sea's temperature; inland, rock and soil with small $c$ heat and cool quickly.
8 $\tfrac12\times\tfrac12mv^2=mc\Delta T$, so $\Delta T=\dfrac{v^2}{4c}=\dfrac{300^2}{4\times128}=1.8\times10^2\ \text{K}$ ($176$). The mass cancels.
II
Calorimetry without change of state.
  • 1 $0.20\ \text{kg}$ of water at $80\ ^\circ\text{C}$ is mixed with $0.30\ \text{kg}$ of water at $20\ ^\circ\text{C}$. Find the final temperature.
  • 2 A $0.500\ \text{kg}$ iron bar at $200\ ^\circ\text{C}$ is plunged into $2.00\ \text{kg}$ of water at $20\ ^\circ\text{C}$. Find the final temperature (ignore the container and any boiling).
  • 3 $0.150\ \text{kg}$ of a metal at $95.0\ ^\circ\text{C}$ is dropped into $0.200\ \text{kg}$ of water at $22.0\ ^\circ\text{C}$ in an $80.0\ \text{g}$ aluminium calorimeter. The final temperature is $27.0\ ^\circ\text{C}$. Find the specific heat capacity of the metal.
  • 4 Find the water equivalent of a copper calorimeter of mass $0.12\ \text{kg}$.
  • 5 How much water at $10\ ^\circ\text{C}$ must be added to $1.0\ \text{kg}$ of water at $70\ ^\circ\text{C}$ to obtain water at $30\ ^\circ\text{C}$?
  • 6 (Modelling) A blacksmith quenches a $1.5\ \text{kg}$ steel horseshoe at $600\ ^\circ\text{C}$ in a bucket containing $8.0\ \text{kg}$ of water at $25\ ^\circ\text{C}$. Estimate the final temperature, ignoring the water that boils away where it touches the hot metal. Will the true final temperature be higher or lower?
  • 7 Why is a calorimeter usually made of thin, polished copper, fitted with a stirrer and a lid and placed inside an insulating jacket?
1 Same substance, so $c$ cancels: $T_f=\dfrac{0.20\times80+0.30\times20}{0.50}=44\ ^\circ\text{C}$.
2 $T_f=\dfrac{0.500\times450\times200+2.00\times4186\times20}{0.500\times450+2.00\times4186}=\dfrac{212\,440}{8597}=24.7\ ^\circ\text{C}$.
3 Gained: $(0.200\times4186+0.0800\times900)\times5.0=4.55\times10^3\ \text{J}$. Lost: $0.150\,c\times68.0$. So $c=\dfrac{4546}{10.2}=4.5\times10^2\ \text{J kg}^{-1}\text{K}^{-1}$ ($446$).
4 $W=\dfrac{0.12\times387}{4186}=0.011\ \text{kg}=11\ \text{g}$.
5 $m(30-10)=1.0(70-30)$, so $m=2.0\ \text{kg}$.
6 $T_f=\dfrac{1.5\times450\times600+8.0\times4186\times25}{1.5\times450+8.0\times4186}=\dfrac{1.242\times10^6}{3.42\times10^4}=36\ ^\circ\text{C}$. Lower in reality: the water that boils takes away latent heat as steam.
7 Thin copper: small heat capacity and good conduction, so the vessel quickly shares the temperature of its contents and takes little heat. Polished: little radiation. Stirrer: uniform temperature. Lid and insulating jacket: little heat lost by convection and conduction — the system is as nearly isolated as possible.
III
Change of state and latent heat.
  • 1 How much heat is needed to melt $2.0\ \text{kg}$ of ice at $0\ ^\circ\text{C}$?
  • 2 How much heat is needed to boil away $0.50\ \text{kg}$ of water already at $100\ ^\circ\text{C}$?
  • 3 How much heat turns $1.0\ \text{kg}$ of ice at $-20\ ^\circ\text{C}$ into water at $30\ ^\circ\text{C}$?
  • 4 Compare the heat given to the skin by $10\ \text{g}$ of steam at $100\ ^\circ\text{C}$ and by $10\ \text{g}$ of water at $100\ ^\circ\text{C}$, each ending as water at $37\ ^\circ\text{C}$.
  • 5 A freezer removes heat at $150\ \text{W}$. How long does it take to turn $1.0\ \text{kg}$ of water at $25\ ^\circ\text{C}$ into ice at $-5\ ^\circ\text{C}$?
  • 6 Explain why food cooks faster in a pressure cooker, and why it is hard to cook rice in an open pot high in the mountains.
  • 7 A wire carrying heavy weights passes slowly through a block of ice, yet the block is still in one piece afterwards. Explain.
  • 8 A $500\ \text{W}$ heater warms $0.40\ \text{kg}$ of a solid. Its temperature rises from $20\ ^\circ\text{C}$ to $80\ ^\circ\text{C}$ in $60\ \text{s}$, then stays at $80\ ^\circ\text{C}$ for $4.0\ \text{min}$ before rising again. Find the specific heat capacity of the solid and its latent heat of fusion (assume no losses).
1 $Q=mL_f=2.0\times3.33\times10^5=6.7\times10^5\ \text{J}$.
2 $Q=mL_v=0.50\times2.26\times10^6=1.13\times10^6\ \text{J}$.
3 $1.0\times2100\times20+1.0\times3.33\times10^5+1.0\times4186\times30=42\,000+333\,000+125\,580=5.0\times10^5\ \text{J}$.
4 Water: $0.010\times4186\times63=2.64\times10^3\ \text{J}$. Steam: $0.010\times2.26\times10^6+2.64\times10^3=2.52\times10^4\ \text{J}$ — about $9.6$ times as much.
5 $Q=1.0\times4186\times25+3.33\times10^5+1.0\times2100\times5=4.48\times10^5\ \text{J}$; $t=\dfrac{4.48\times10^5}{150}=2.99\times10^3\ \text{s}\approx50\ \text{min}$.
6 The boiling point rises with pressure: inside the sealed cooker water boils at about $120\ ^\circ\text{C}$, and cooking reactions go much faster at the higher temperature. On a mountain the air pressure is lower, water boils below $100\ ^\circ\text{C}$ (near $88\ ^\circ\text{C}$ at $3500\ \text{m}$), and food cooks slowly however hard it boils.
7 Regelation: the large pressure under the thin wire lowers the melting point of the ice below its temperature, so the ice there melts; the water flows round to the top of the wire, where the pressure is normal again, and refreezes, giving up its latent heat — which helps to melt more ice below the wire.
8 Solid: $500\times60=0.40\,c\times60$, so $c=1.25\times10^3\ \text{J kg}^{-1}\text{K}^{-1}$. Plateau: $L_f=\dfrac{500\times240}{0.40}=3.0\times10^5\ \text{J/kg}$; it melts at $80\ ^\circ\text{C}$.
IV
Mixtures with a change of state.
  • 1 $50\ \text{g}$ of ice at $0\ ^\circ\text{C}$ is dropped into $200\ \text{g}$ of water at $40\ ^\circ\text{C}$. Find the final temperature.
  • 2 $0.50\ \text{kg}$ of ice at $0\ ^\circ\text{C}$ is mixed with $0.50\ \text{kg}$ of water at $40\ ^\circ\text{C}$. Find the final temperature and the mass of ice remaining.
  • 3 $20\ \text{g}$ of steam at $100\ ^\circ\text{C}$ is bubbled into $0.50\ \text{kg}$ of water at $20\ ^\circ\text{C}$ until it has all condensed. Find the final temperature.
  • 4 What is the least mass of ice at $0\ ^\circ\text{C}$ that will cool $0.25\ \text{kg}$ of a juice (treat it as water) from $30\ ^\circ\text{C}$ to $0\ ^\circ\text{C}$?
  • 5 (Modelling) A marathon runner on a hot day loses about $0.50\ \text{kg}$ of sweat per hour, all of which evaporates. Taking $L_v=2.4\times10^6\ \text{J/kg}$ at skin temperature, find the rate at which evaporation removes heat from the body.
1 $Q_\text{avail}=0.200\times4186\times40=3.35\times10^4\ \text{J}$ exceeds $Q_\text{need}=0.050\times3.33\times10^5=1.67\times10^4\ \text{J}$: all melts. Then $0.200\times4186(40-T)=16\,650+0.050\times4186\,T$, giving $T=\dfrac{16\,838}{1046.5}=16\ ^\circ\text{C}$ ($16.1$).
2 $Q_\text{avail}=0.50\times4186\times40=8.37\times10^4\ \text{J}$, less than $0.50\times3.33\times10^5=1.67\times10^5\ \text{J}$: final temperature $0\ ^\circ\text{C}$. Melted $=\dfrac{8.37\times10^4}{3.33\times10^5}=0.25\ \text{kg}$, so $0.25\ \text{kg}$ of ice remains.
3 Steam condenses and cools: $0.020\times2.26\times10^6+0.020\times4186(100-T)=0.50\times4186(T-20)$. So $45\,200+8372-83.7T=2093T-41\,860$, $T=\dfrac{95\,432}{2176.7}=43.8\ ^\circ\text{C}$. (Check that all the steam condenses: the water could absorb $0.50\times4186\times80=1.67\times10^5\ \text{J}$ before reaching $100\ ^\circ\text{C}$, far more than the $4.52\times10^4\ \text{J}$ released on condensing.)
4 $mL_f=0.25\times4186\times30$, so $m=\dfrac{31\,395}{3.33\times10^5}=0.094\ \text{kg}=94\ \text{g}$ (all of it ending as water at $0\ ^\circ\text{C}$).
5 $P=\dfrac{0.50\times2.4\times10^6}{3600}=3.3\times10^2\ \text{W}$ — comparable with the heat a hard-working body must get rid of.
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 2 marks Define specific heat capacity and latent heat of fusion, giving their SI units. Why does the temperature of melting ice not rise although heat is being supplied?

Definitions Specific heat capacity: heat needed to raise the temperature of unit mass of a substance by one kelvin, $c=\dfrac{Q}{m\Delta T}$, unit $\text{J kg}^{-1}\text{K}^{-1}$. Latent heat of fusion: heat needed to change unit mass of the solid into liquid at its melting point without change of temperature, $L_f=\dfrac Qm$, unit $\text{J/kg}$. [1]
Reason During melting the heat supplied is used to break the ordered arrangement of the molecules (doing work against their attractions), not to increase their average kinetic energy, so the temperature stays at $0\ ^\circ\text{C}$ until all the ice has melted. [1]
Exam tip Always state "at constant temperature" in the definition of latent heat.

E2. NEET Taking the specific heat capacity of water as $1\ \text{cal g}^{-1}\,{}^\circ\text{C}^{-1}$, $L_f=80\ \text{cal/g}$ and $L_v=540\ \text{cal/g}$, the heat needed to convert $1\ \text{g}$ of ice at $0\ ^\circ\text{C}$ into steam at $100\ ^\circ\text{C}$ is (a) $620\ \text{cal}$ (b) $640\ \text{cal}$ (c) $720\ \text{cal}$ (d) $180\ \text{cal}$

Steps Melt: $80$; heat $0\to100\ ^\circ\text{C}$: $1\times1\times100=100$; boil: $540$. Total $720\ \text{cal}$.
Answer (c).
Exam tip (a) forgets heating the water, (b) forgets melting, (d) forgets boiling. Write one term per stage of the heating curve.

E3. JEE Main (Numerical value) $100\ \text{g}$ of ice at $-10\ ^\circ\text{C}$ is put into $200\ \text{g}$ of water at $30\ ^\circ\text{C}$ in an insulated container of negligible heat capacity. Using $c_\text{ice}=2100$, $c_\text{w}=4186\ \text{J kg}^{-1}\text{K}^{-1}$ and $L_f=3.33\times10^5\ \text{J/kg}$, find the mass of ice, in grams, left when equilibrium is reached (nearest integer).

Available Water cooling to $0\ ^\circ\text{C}$: $0.200\times4186\times30=25\,116\ \text{J}$.
Needed Warm the ice to $0\ ^\circ\text{C}$: $0.100\times2100\times10=2100\ \text{J}$; melt all of it: $33\,300\ \text{J}$; total $35\,400\ \text{J}\gt25\,116\ \text{J}$, so the final temperature is $0\ ^\circ\text{C}$ and only part melts.
Melted $\dfrac{25\,116-2100}{3.33\times10^5}=0.0691\ \text{kg}$; left $100-69.1=30.9\ \text{g}$.
Answer $31$. Exam tip: the heat to warm the ice to $0\ ^\circ\text{C}$ comes out of the available heat before any melting.

E4. JEE Advanced (One or more correct) A $0.50\ \text{kg}$ sample is heated by a $400\ \text{W}$ heater with no heat losses. Its temperature rises uniformly from $-30\ ^\circ\text{C}$ to $0\ ^\circ\text{C}$ in the first $60\ \text{s}$, stays at $0\ ^\circ\text{C}$ for the next $300\ \text{s}$, and then rises uniformly to $40\ ^\circ\text{C}$ in the next $120\ \text{s}$. Which statements are correct? (A) The specific heat capacity of the solid is $1.6\times10^3\ \text{J kg}^{-1}\text{K}^{-1}$. (B) The latent heat of fusion is $2.4\times10^5\ \text{J/kg}$. (C) The specific heat capacity of the liquid is smaller than that of the solid. (D) On a temperature–time graph the liquid stage has a smaller slope than the solid stage.

(A) $c_s=\dfrac{Pt}{m\Delta T}=\dfrac{400\times60}{0.50\times30}=1600\ \text{J kg}^{-1}\text{K}^{-1}$ ✓.
(B) $L_f=\dfrac{Pt}{m}=\dfrac{400\times300}{0.50}=2.4\times10^5\ \text{J/kg}$ ✓.
(C) $c_l=\dfrac{400\times120}{0.50\times40}=2400\ \text{J kg}^{-1}\text{K}^{-1}$, larger than $c_s$ ✗.
(D) Slopes: solid $\dfrac{30}{60}=0.50\ \text{K/s}$, liquid $\dfrac{40}{120}=0.33\ \text{K/s}$ ✓ — at constant power, slope $=\dfrac{P}{mc}$, so the larger $c$ gives the gentler slope.
Answer (A), (B), (D).
Exam tip On a constant-power heating curve, read $c$ from the slopes and $L$ from the length of a plateau — exactly as in Interactive 2.

E5. JEE Main NEET Two solid spheres of the same material have radii $r$ and $2r$. They are given equal amounts of heat. The ratio of the rise in temperature of the smaller sphere to that of the larger is (a) $1:2$ (b) $2:1$ (c) $4:1$ (d) $8:1$

Idea $\Delta T=\dfrac{Q}{mc}$ with $m\propto r^3$: the larger sphere has $8$ times the mass.
Answer $\Delta T_1:\Delta T_2=8:1$: (d).
Exam tip (c) uses surface area ($r^2$) instead of volume; heat capacity goes with mass.

E6. Board · Case study A family in a hill town finds that dal cooked in an open pot never softens, so they buy a pressure cooker. Its valve keeps the pressure inside at about twice atmospheric pressure, where water boils at about $120\ ^\circ\text{C}$. (i) Why does food cook faster in the pressure cooker? [1] (ii) Why does water boil below $100\ ^\circ\text{C}$ in the hill town? [1] (iii) The cooker holds $1.5\ \text{kg}$ of water at $25\ ^\circ\text{C}$. Find the heat needed to bring it to $120\ ^\circ\text{C}$ and then to turn $50\ \text{g}$ of it into steam (take $L_v=2.2\times10^6\ \text{J/kg}$ at $120\ ^\circ\text{C}$). [2]

(i) The higher pressure raises the boiling point, so the water and food reach about $120\ ^\circ\text{C}$ instead of $100\ ^\circ\text{C}$; cooking (a set of chemical reactions) is much faster at the higher temperature. [1]
(ii) Atmospheric pressure is lower at altitude, and the boiling point falls with pressure. [1]
(iii) Heating: $1.5\times4186\times95=5.97\times10^5\ \text{J}$. Steam: $0.050\times2.2\times10^6=1.1\times10^5\ \text{J}$. Total $7.1\times10^5\ \text{J}$. [2]
Exam tip $L_v$ itself falls slightly as the boiling temperature rises, which is why a different value is given at $120\ ^\circ\text{C}$ — use the value the question supplies.
🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. How long can an insulated box packed with ice keep vaccines between $2\ ^\circ\text{C}$ and $8\ ^\circ\text{C}$ on the way to a remote village?

Why it matters. Many vaccines lose their potency if they get too warm — or if they freeze. Immunisation campaigns in India, sub-Saharan Africa and South-East Asia rely on a "cold chain" whose last link is often a vaccine carrier with frozen ice packs, carried on foot, by motorbike or by boat for hours or days. The latent heat of fusion of the ice is what buys that time.

What to do. Model a carrier holding four ice packs, each with $0.40\ \text{kg}$ of water frozen at about $0\ ^\circ\text{C}$. Its insulated walls let heat leak in from the warm surroundings at a steady rate $P_\text{leak}$ (Section 10.3 shows how to estimate it).

  1. Explain why, while any ice remains, the inside of the box stays close to $0\ ^\circ\text{C}$.
  2. With $P_\text{leak}=5.0\ \text{W}$, find the time for all the ice to melt.
  3. Packs taken straight from a freezer at $-20\ ^\circ\text{C}$ can freeze the vaccines. How much extra cooling capacity do they carry, as a percentage of the latent heat? Why do health workers "condition" packs (let them warm until they start to melt) before use?
  4. Find out the typical "cold life" claimed for real vaccine carriers at an outside temperature of $43\ ^\circ\text{C}$, and use your model to work out the leak rate it implies.
  5. Suggest two design changes that would double the cold life, and estimate their cost in mass or volume.

Tools: Calculator or spreadsheet; published performance specifications of vaccine carriers and cold boxes from international health agencies; a thermometer, an ice-box and ice for a home test.

Hand in: Your calculations, a graph of temperature inside the box against time (flat while ice remains, rising afterwards), and a half-page recommendation for a health worker.

Going further. "Phase-change materials" with melting points near $5\ ^\circ\text{C}$ (salt solutions, paraffins) can hold a box at a safer temperature than ice. Find one, look up its latent heat, and compare.

Instructor note: step 2: $1.6\times3.33\times10^5/5.0=1.07\times10^5\ \text{s}\approx30\ \text{h}$. Step 3: warming the ice from $-20$ to $0\ ^\circ\text{C}$ absorbs $2100\times20=4.2\times10^4\ \text{J/kg}$, about $13\%$ of $L_f$; the danger is not extra time but sub-zero temperatures near the vaccines. Step 4: a claimed cold life of, say, $36\ \text{h}$ implies $P_\text{leak}\approx4\ \text{W}$. Doubling the wall thickness roughly halves the leak (Section 10.3).

🧪 Self-check quiz

0/5 correct

Q1. The SI unit of specific heat capacity is

Q2. While ice at $0\ ^\circ\text{C}$ is melting on a steady flame, its temperature

Q3. Equal masses of water at $60\ ^\circ\text{C}$ and $20\ ^\circ\text{C}$ are mixed in an insulated flask. The final temperature is

Q4. For water, the latent heat of vaporisation is about

Q5. $100\ \text{g}$ of ice at $0\ ^\circ\text{C}$ is mixed with $100\ \text{g}$ of water at $80\ ^\circ\text{C}$ ($L_f=80\ \text{cal/g}$, $c_w=1\ \text{cal g}^{-1}\,{}^\circ\text{C}^{-1}$). The final temperature is

← 10.1 Temperature, the Ideal-Gas Law and Thermal Expansion10.3 Heat Transfer and Newton's Law of Cooling →