Specific Heat Capacity and Specific Latent Heat

A2 · 13 min

Supplying energy to a substance does one of two things: it raises the temperature, or it changes the state (melting or boiling) at constant temperature. Specific heat capacity measures how much energy the first takes; specific latent heat measures the second. Both are defined word for word in Paper 4, both appear in calculations every year, and the electrical methods for measuring them are classic Paper 5 planning and evaluation contexts.

Specific heat capacity

Heating $1\ \text{kg}$ of water by $1\ \text{K}$ takes about $4200\ \text{J}$ ; heating $1\ \text{kg}$ of copper by $1\ \text{K}$ takes only about $390\ \text{J}$ . The energy needed depends on the mass, the temperature rise and the material.

Definition

The specific heat capacity $c$ of a substance is the energy required per unit mass of the substance to raise its temperature by one kelvin (or one degree Celsius).

Key result

Q = mc\Delta\theta

$Q$ : energy transferred (J)
$m$ : mass (kg)
$c$ : specific heat capacity ( $\text{J kg}^{-1}\ \text{K}^{-1}$ )
$\Delta\theta$ : temperature change (K or $^\circ$ C; the same number)

Because $\Delta\theta$ is a temperature change, it has the same value in kelvin and in degrees Celsius. The unit of $c$ can be written $\text{J kg}^{-1}\ \text{K}^{-1}$ or $\text{J kg}^{-1}\ ^\circ\text{C}^{-1}$ .

What the energy does

When a substance is heated without changing state, the energy increases the random kinetic energy of its molecules (and, in solids and liquids, also their potential energy as they vibrate further from their equilibrium positions). The temperature rises because temperature is related to the mean kinetic energy of the molecules. Water has a high specific heat capacity, so it stores a lot of energy for a small temperature rise, which is why it is used in central heating and car cooling systems.

Heating with a constant power

If an electrical heater of power $P$ supplies energy for a time $t$ , then $Q = Pt$ (or $Q = VIt$ ). With no energy losses,

Pt = mc\Delta\theta \quad\Rightarrow\quad \frac{\Delta\theta}{t} = \frac{P}{mc}

So the gradient of a graph of temperature against time is $P/(mc)$ , and $c = P/(m \times \text{gradient})$ . Using the gradient of a graph rather than a single pair of readings averages out reading errors and lets you use the straight central portion, avoiding the start (where the heater itself is warming up) and the end (where losses are largest).

Specific latent heat

When ice at $0\ ^\circ\text{C}$ is heated, it melts, but its temperature stays at $0\ ^\circ\text{C}$ until it has all melted. The energy supplied is used to change the state, not to raise the temperature. This "hidden" energy is called latent heat.

Definition

The specific latent heat $L$ of a substance is the energy required per unit mass of the substance to change its state without any change in temperature.

The specific latent heat of fusion is the energy required per unit mass to change a solid into a liquid without any change in temperature.
The specific latent heat of vaporisation is the energy required per unit mass to change a liquid into a gas (vapour) without any change in temperature.

Key result

Q = mL

Unit of $L$ : $\text{J kg}^{-1}$ . For water: $L_f = 3.34 \times 10^{5}\ \text{J kg}^{-1}$ and $L_v = 2.26 \times 10^{6}\ \text{J kg}^{-1}$ .

The same energy is released when the change is reversed: condensing $1\ \text{kg}$ of steam releases $2.26 \times 10^{6}\ \text{J}$ , which is why scalds from steam are so much worse than from boiling water.

What happens to the molecules

During a change of state the energy supplied increases the potential energy of the molecules by breaking or weakening the bonds between them. The mean kinetic energy of the molecules stays the same, so the temperature does not change.

Why vaporisation needs much more energy than fusion

For water, $L_v$ is almost seven times $L_f$ . Two reasons:

Separation. On melting, the molecules stay close together (the density barely changes); only some bonds are broken. On boiling, the molecules are separated completely, to distances about ten times greater: every bond is broken, so the increase in potential energy is far larger.
Work done on the atmosphere. A gas occupies a much larger volume than the liquid. As the vapour forms, it must push back the surrounding atmosphere, doing work $p\Delta V$ . On melting, the volume change is tiny, so this work is negligible.

Heating curve

The graph shows how the temperature of a substance changes when it is heated at constant power, starting as a solid below its melting point.

(0, -20) -- (2, 0) (2, 0) -- (6, 0) (6, 0) -- (10, 100) (10, 100) -- (28, 100) (28, 100) -- (30, 120)

Reading the curve (time in minutes, temperature in $^\circ$ C for water):

Sloping sections: one state is being heated. Energy raises the kinetic energy of molecules; temperature rises. The gradient is $P/(mc)$ , so a steeper slope means a smaller specific heat capacity. Ice ( $c = 2100\ \text{J kg}^{-1}\ \text{K}^{-1}$ ) warms about twice as fast as liquid water ( $4200$ ).
Flat sections: change of state. Energy increases potential energy; temperature constant. The length of each flat section is proportional to the latent heat, so the boiling plateau is much longer than the melting plateau.

Measuring specific heat capacity: the electrical method

Specific heat capacity of a metal block

Apparatus: metal block of known mass (measured on a balance) with two holes drilled in it; an electrical immersion heater in one hole and a thermometer (or temperature sensor) in the other, with a drop of oil in each hole for good thermal contact; insulating jacket (expanded polystyrene or lagging) around the block; power supply, ammeter, voltmeter (or a joulemeter); stopwatch.

Method:

Measure the mass $m$ of the block.
Record the starting temperature, then switch on the heater and start the stopwatch.
Record $V$ and $I$ (check they stay constant) and record the temperature every $30\ \text{s}$ for about $10$ minutes.
Plot temperature $\theta$ against time $t$ and find the gradient of the straight section.
Calculate $c = \dfrac{VI}{m \times \text{gradient}}$ .

Variables: independent variable is time; dependent variable is temperature; control variables are the heater power and the mass of the block.

Sources of error and improvements:

Energy is lost to the surroundings, so the temperature rise is smaller than expected and the measured $c$ is too large. Improve: insulate the block; start below room temperature and finish the same amount above it so gains and losses roughly cancel.
The heater and thermometer take time to warm up, so the temperature lags behind the energy supplied (the graph curves at the start). Use the gradient of the straight central section.
After switching off, the temperature keeps rising briefly as energy reaches the thermometer. Record the maximum temperature reached.
Poor thermal contact between thermometer and block: use oil in the holes.

For liquids: place the liquid in an insulated container (with a lid), stir continuously, and include the energy absorbed by the container, or use a container with negligible heat capacity.

Measuring specific latent heat

Specific latent heat of vaporisation of water

Apparatus: beaker or kettle of water on a top-pan balance; immersion heater connected through a joulemeter (or with an ammeter and voltmeter); insulation around the beaker; stopwatch.

Method:

Heat the water until it is boiling steadily.
Record the balance reading and the joulemeter reading (or start the stopwatch), then let the water boil for a measured time, for example $5$ minutes.
Record the new balance reading and energy supplied. The mass boiled away is $\Delta m$ and $L_v = E/\Delta m$ .

Eliminating heat losses: repeat at a different heater power. If the rate of energy loss $h$ is the same both times (same temperature, same apparatus):

P_1 t - ht = m_1L_v, \qquad P_2t - ht = m_2L_v \quad\Rightarrow\quad L_v = \frac{(P_2 - P_1)t}{m_2 - m_1}

Errors: some vapour condenses and drips back, giving a smaller apparent $\Delta m$ ; water may splash out; the heater must be fully immersed.

Specific latent heat of fusion of ice

Two identical funnels are filled with crushed ice at $0\ ^\circ\text{C}$ ; one contains an immersion heater. Water dripping from each is collected in beakers over the same time. The control funnel (no heater) measures the ice melted by energy from the surroundings. Then

L_f = \frac{Pt}{m_{\text{heater}} - m_{\text{control}}}

Use crushed ice so the heater is in good contact; dry the ice first, or the collected water will include water that was never ice.

Thermal energy calculations

List every stage: warming a solid, melting, warming a liquid, boiling, and so on.
For each stage write $Q = mc\Delta\theta$ or $Q = mL$ , with its own $c$ or $L$ .
Add the stages for the total energy.
For mixing problems: energy lost by the hot object = energy gained by the cold object (assuming no losses). Write the unknown final temperature as $T$ in every term.
For electrical heating, use $Q = Pt = VIt$ .

Worked examples

Heating water

How much energy is needed to heat $0.50\ \text{kg}$ of water from $20\ ^\circ\text{C}$ to $80\ ^\circ\text{C}$ ? ( $c = 4200\ \text{J kg}^{-1}\ \text{K}^{-1}$ )

Solution

Q = mc\Delta\theta = 0.50 \times 4200 \times 60 = 1.26 \times 10^{5}\ \text{J}

Electrical method with a graph

A $1.0\ \text{kg}$ aluminium block is heated by a $50\ \text{W}$ heater. Its temperature rises from $20.0\ ^\circ\text{C}$ to $31.5\ ^\circ\text{C}$ in $4.0$ minutes. (a) Calculate the specific heat capacity indicated. (b) The accepted value is $900\ \text{J kg}^{-1}\ \text{K}^{-1}$ . Explain the difference.

Solution

(a)

c = \frac{Pt}{m\Delta\theta} = \frac{50 \times 240}{1.0 \times 11.5} = 1.04 \times 10^{3}\ \text{J kg}^{-1}\ \text{K}^{-1}

(b) Some of the energy supplied is lost to the surroundings, so less than $Pt$ actually heats the block. The temperature rise is smaller than it would be without losses, so the calculated $c$ is too large.

From ice to warm water

Calculate the energy needed to turn $0.20\ \text{kg}$ of ice at $-10\ ^\circ\text{C}$ into water at $25\ ^\circ\text{C}$ . ( $c_{\text{ice}} = 2100\ \text{J kg}^{-1}\ \text{K}^{-1}$ , $c_{\text{water}} = 4200\ \text{J kg}^{-1}\ \text{K}^{-1}$ , $L_f = 3.34 \times 10^{5}\ \text{J kg}^{-1}$ )

Solution

Warm the ice to $0\ ^\circ\text{C}$ : $0.20 \times 2100 \times 10 = 4200\ \text{J}$

Melt it: $0.20 \times 3.34 \times 10^{5} = 66\,800\ \text{J}$

Warm the water to $25\ ^\circ\text{C}$ : $0.20 \times 4200 \times 25 = 21\,000\ \text{J}$

Total: $4200 + 66\,800 + 21\,000 = 9.2 \times 10^{4}\ \text{J}$ .

Notice that melting takes far more energy than both warming stages combined.

Ice in a drink

An ice cube of mass $0.050\ \text{kg}$ at $0\ ^\circ\text{C}$ is dropped into $0.30\ \text{kg}$ of water at $30\ ^\circ\text{C}$ in an insulated cup of negligible heat capacity. Calculate the final temperature.

Solution

Let the final temperature be $T$ (in $^\circ$ C). Energy lost by the warm water = energy gained by the ice (to melt, then to warm from $0$ to $T$ ):

0.30 \times 4200 \times (30 - T) = 0.050 \times 3.34 \times 10^{5} + 0.050 \times 4200 \times T

37\,800 - 1260T = 16\,700 + 210T \quad\Rightarrow\quad 1470T = 21\,100 \quad\Rightarrow\quad T = 14\ ^\circ\text{C}

Check that the answer makes sense: it lies between $0$ and $30\ ^\circ\text{C}$ , and all the ice has melted (the warm water could supply up to $37\,800\ \text{J}$ , more than the $16\,700\ \text{J}$ needed to melt the ice).

Latent heat with heat losses eliminated

Water is kept boiling by an immersion heater. With the heater at $60.0\ \text{W}$ , $6.90\ \text{g}$ of water boils away in $300\ \text{s}$ . At $100.0\ \text{W}$ , $12.21\ \text{g}$ boils away in $300\ \text{s}$ . Calculate the specific latent heat of vaporisation of water and the rate of energy loss to the surroundings.

Solution

The rate of energy loss $h$ is the same in both runs:

L_v = \frac{(P_2 - P_1)t}{m_2 - m_1} = \frac{(100.0 - 60.0) \times 300}{(12.21 - 6.90) \times 10^{-3}} = \frac{12\,000}{5.31 \times 10^{-3}} = 2.26 \times 10^{6}\ \text{J kg}^{-1}

Energy loss: $P_1t - ht = m_1L_v$ , so

h = P_1 - \frac{m_1L_v}{t} = 60.0 - \frac{6.90 \times 10^{-3} \times 2.26 \times 10^{6}}{300} = 60.0 - 52.0 = 8.0\ \text{W}

Watch out

Using $mc\Delta\theta$ during a change of state. There is no temperature change while melting or boiling, so $mc\Delta\theta = 0$ for that stage. Use $mL$ .

Watch out

Saying that temperature stays constant during melting "because no energy is supplied". Energy is supplied; it increases the potential energy of the molecules (breaking bonds), not their kinetic energy.

Watch out

Defining latent heat without "without change of temperature". The definition must include that the change of state happens at constant temperature, and it must be per unit mass for the specific quantity.

Exam tip

"Define specific latent heat of fusion" (2 marks): energy per unit mass; to change solid to liquid; without change of temperature. Each part matters.
"Explain why $L_v > L_f$ " (2–3 marks): greater increase in molecular separation on boiling (all bonds broken, greater increase in potential energy) and work is done against the atmosphere as the gas expands.
In "explain why the experimental value is too high/low" questions, state the direction of the effect and why: "energy is lost to the surroundings, so the temperature rise is less than expected, so $c$ calculated is larger than the true value".
In mixing calculations, write the energy balance equation in words first; the examiner awards a method mark for it.

Summary

Specific heat capacity: energy per unit mass to raise the temperature by one kelvin; $Q = mc\Delta\theta$ .
Specific latent heat: energy per unit mass to change state without change of temperature; $Q = mL$ .
Fusion: solid to liquid; vaporisation: liquid to gas. For water $L_v \gg L_f$ .
During heating, energy increases molecular kinetic energy; during a change of state, it increases molecular potential energy.
On a heating curve at constant power, slopes are $P/(mc)$ and flat sections are changes of state.
Electrical methods: heat losses make measured $c$ and $L$ too large; use insulation, graph gradients, or two powers to cancel losses.

Practice questions

Question

Define specific heat capacity.
A $9.0\ \text{kW}$ electric shower heats water from $15\ ^\circ\text{C}$ to $40\ ^\circ\text{C}$ . Calculate the maximum mass of water it can heat per second.
Calculate the energy released when $5.0\ \text{g}$ of steam at $100\ ^\circ\text{C}$ condenses and then cools to $37\ ^\circ\text{C}$ . Compare with the energy released by $5.0\ \text{g}$ of water at $100\ ^\circ\text{C}$ cooling to $37\ ^\circ\text{C}$ .
A $40\ \text{W}$ heater warms $0.80\ \text{kg}$ of a liquid; the graph of temperature against time has gradient $0.050\ \text{K s}^{-1}$ . Calculate the specific heat capacity of the liquid, assuming no losses.
A lead bullet of mass $10\ \text{g}$ travelling at $300\ \text{m s}^{-1}$ hits a wall and stops. Assuming all its kinetic energy heats the bullet, calculate its temperature rise. ( $c_{\text{lead}} = 130\ \text{J kg}^{-1}\ \text{K}^{-1}$ )
Explain, in terms of molecules, why the temperature of a melting solid does not change even though energy is being supplied.
A freezer removes energy at $100\ \text{W}$ from $0.50\ \text{kg}$ of water at $20\ ^\circ\text{C}$ . Calculate the minimum time to turn it into ice at $-5\ ^\circ\text{C}$ .
In a funnel experiment, a $24\ \text{W}$ heater runs for $600\ \text{s}$ . The heated funnel collects $52.0\ \text{g}$ of water and the control funnel $9.0\ \text{g}$ . (a) Calculate the specific latent heat of fusion of ice. (b) Explain the purpose of the control funnel. (c) Suggest why the result might still be inaccurate.

Answers

The energy required per unit mass of a substance to raise its temperature by one kelvin.
$m/t = P/(c\Delta\theta) = 9000/(4200 \times 25) = 0.086\ \text{kg s}^{-1}$ .
Steam: $0.0050 \times 2.26 \times 10^{6} + 0.0050 \times 4200 \times 63 = 11\,300 + 1323 = 1.26 \times 10^{4}\ \text{J}$ . Water only: $1323\ \text{J} \approx 1.3 \times 10^{3}\ \text{J}$ , nearly ten times less.
$c = P/(m \times \text{gradient}) = 40/(0.80 \times 0.050) = 1.0 \times 10^{3}\ \text{J kg}^{-1}\ \text{K}^{-1}$ .
$E_K = \tfrac{1}{2} \times 0.010 \times 300^2 = 450\ \text{J}$ ; $\Delta\theta = 450/(0.010 \times 130) = 350\ \text{K}$ (enough to start melting the lead).
The energy supplied increases the potential energy of the molecules, breaking bonds between them; the mean kinetic energy of the molecules (which determines temperature) does not change.
$E = 0.50 \times 4200 \times 20 + 0.50 \times 3.34 \times 10^{5} + 0.50 \times 2100 \times 5 = 42\,000 + 167\,000 + 5250 = 2.14 \times 10^{5}\ \text{J}$ ; $t = E/P = 2.14 \times 10^{3}\ \text{s}$ (about $36$ minutes).
(a) $L_f = Pt/(m_h - m_c) = 24 \times 600/(0.0430) = 3.3 \times 10^{5}\ \text{J kg}^{-1}$ . (b) It measures the ice melted by energy from the surroundings in the same time; subtracting it leaves the mass melted by the heater alone. (c) Water may remain in the ice rather than drip through; the ice may not have been at exactly $0\ ^\circ\text{C}$ or may have been wet; the two funnels may not receive equal energy from the surroundings (the heated funnel's ice is in contact with a warm heater).