Quantity of Heat and Heat Transfer

Lessons

Senior 3 · Physics

Quantity of Heat and Heat Transfer

5 lessons

Lesson 2 of 5

The Principle of Heat Exchange and the Method of Mixtures

25 min

Getting the bath water right

Before bathing a baby, a parent pours water heated on the stove into a basin of cold water from the jerrycan, a little at a time, testing with an elbow until it feels warm but not hot. The hot water cools down and the cold water warms up until both reach the same temperature.

Nothing mysterious happens in the basin: heat simply moves from the hot water to the cold water. If we know how much heat each part can give or take, we can predict the final temperature before we pour, or work out how much hot water we need. This lesson turns that idea into a method that is also used in the laboratory to measure specific heat capacities.

The principle of heat exchange

When a hot body and a cold body are put in contact, heat flows from the hot one to the cold one until they reach the same temperature (thermal equilibrium, which you met in S1). Energy is conserved, so if no heat escapes to the surroundings, every joule lost by the hot body is gained by the cold body:

\text{heat lost by the hot bodies} = \text{heat gained by the cold bodies}

If a hot body at $T_{h}$ and a cold body at $T_{c}$ reach a common final temperature $T$ , then

m_h\,c_h\,(T_h - T) = m_c\,c_c\,(T - T_c)

Each bracket is written so that it is positive: hot minus final for the body that cools, final minus cold for the body that warms.

Definition

Principle of heat exchange (principle of mixtures): when bodies at different temperatures are mixed or put in contact, and no heat is exchanged with the surroundings, the heat lost by the hotter bodies equals the heat gained by the colder bodies. Example: hot and cold water mixed in a lagged flask.

Calorimeter: a container, usually a thin copper or aluminium can with a lid, stirrer and insulated outer jacket, in which heat exchanges are measured. The method of mixtures is the use of the principle of heat exchange in a calorimeter to find an unknown specific heat capacity or temperature.

Worked example 1: mixing hot and cold water

4.0 kg of water at 80 °C is mixed with 6.0 kg of water at 20 °C in a plastic basin. Ignoring heat taken by the basin and lost to the air, find the final temperature $T$ .

4.0 \times 4200 \times (80 - T) = 6.0 \times 4200 \times (T - 20)

The 4200 cancels on both sides:

320 - 4.0\,T = 6.0\,T - 120 \quad\Rightarrow\quad 10\,T = 440 \quad\Rightarrow\quad T = 44\ ^\circ\text{C}

Check: 44 °C lies between 20 °C and 80 °C, and nearer to 20 °C because there is more cold water.

Worked example 2: how much cold water?

What mass of cold water at 20 °C must be added to 3.0 kg of water at 90 °C so that the mixture is at 40 °C?

\text{heat lost by hot water} = 3.0 \times 4200 \times (90 - 40) = 630\,000\ \text{J}

\text{heat gained by cold water} = m \times 4200 \times (40 - 20) = 84\,000\,m

84\,000\,m = 630\,000 \quad\Rightarrow\quad m = 7.5\ \text{kg}

Experiment: the specific heat capacity of a metal by the method of mixtures

Find the mass of the empty calorimeter $m_{cal}$ , then its mass with some cold water in it, to get the mass of water $m_{w}$ . Read the water's temperature $T_{1}$ .
Hang the metal block (mass $m$ ) on a thread in boiling water for several minutes, so that it reaches the water's temperature $T_{2}$ . Read $T_{2}$ on the thermometer; do not assume 100 °C (lesson 3 explains why water boils below 100 °C in Kigali).
Move the metal quickly into the calorimeter, close the lid and stir gently.
Record the highest steady temperature $T$ of the mixture.

The metal cools from $T_{2}$ to $T$ ; the water and the calorimeter both warm from $T_{1}$ to $T$ :

m\,c\,(T_2 - T) = \left( m_w\,c_w + m_{cal}\,c_{cal} \right)(T - T_1)

Important

Safety. Boiling water scalds. Keep the beaker on a steady hot plate away from the bench edge, lift the metal by its thread, and move it with the calorimeter held close to the beaker. Wipe up spilt water at once. If anyone is scalded, cool the burn at once under cool (not ice-cold) running water for 20 minutes and tell the teacher; never put ice, butter or toothpaste on it.

Worked example 3: identifying a metal

A 0.20 kg metal block is heated in boiling water at 95 °C and moved into a 0.10 kg copper calorimeter ( $c = 390\ \mathrm{J\,kg^{-1}\,K^{-1}}$ ) holding 0.30 kg of water at 20.0 °C. The final temperature is 24.0 °C. Find the specific heat capacity of the metal.

\text{gained} = (0.30 \times 4200 + 0.10 \times 390)(24.0 - 20.0) = 1299 \times 4.0 = 5196\ \text{J}

\text{lost} = 0.20 \times c \times (95 - 24.0) = 14.2\,c

14.2\,c = 5196 \quad\Rightarrow\quad c \approx 370\ \mathrm{J\,kg^{-1}\,K^{-1}}

This is close to brass (about 380) or copper (390). The result comes out a little low because the block cools slightly during the move, so it really starts below 95 °C.

Warning

Do not forget the container. The calorimeter warms up along with the water, so it gains heat too. Leaving out the $m_{cal}\,c_{cal}$ term makes the calculated $c$ of the metal too small. And always check that your final temperature lies between the starting temperatures; if it does not, a sign is wrong.

Getting a good result

Source of error	Effect	How to reduce it
The metal cools while it is being moved	It starts below the temperature you recorded, so $c$ comes out too low	Move it quickly, with the calorimeter close to the beaker
Heat escapes from the calorimeter	The water warms less, so $c$ comes out too low	Lagging, a lid, and a polished outer surface
Boiling water carried over on the metal	Extra heat enters the calorimeter	Shake the metal once before dropping it in
Uneven temperature in the water	The thermometer reads a local value	Stir gently before every reading

Tip

Lost on the left, gained on the right. Write every body that cools on one side, every body that warms on the other, each as $mc \times$ (bigger temperature minus smaller). The final temperature always lies between the starting ones, closer to the body with the larger $m c$ .

Specific heat capacity of solid and liquid (Amrita Create). Watch the calorimeter experiment done step by step: weighing the calorimeter, heating the solid, transferring it and reading the final temperature.

✦

Practice

State the principle of heat exchange. Under what condition is it exactly true?
2.0 kg of water at 70 °C is mixed with 3.0 kg of water at 20 °C. Find the final temperature, ignoring heat losses.
What mass of water at 15 °C must be added to 1.2 kg of water at 90 °C to give a mixture at 50 °C?
A 0.50 kg aluminium block at 100 °C is dropped into 0.40 kg of water at 20 °C. Ignoring the container and heat losses, find the final temperature.
A 0.10 kg metal at 95 °C is put into 0.20 kg of water at 21.0 °C in a 0.050 kg aluminium calorimeter. The final temperature is 24.5 °C. Find the specific heat capacity of the metal. Which metal could it be?
A blacksmith plunges a 0.80 kg iron hoe blade at 400 °C into 5.0 kg of water at 20 °C. Ignoring steam and heat losses, find the final temperature.
Explain why each of these makes a measured specific heat capacity too low: (a) the metal cools while it is carried to the calorimeter; (b) the calorimeter has no lid.

Show answersHide answers

When bodies at different temperatures exchange heat, the heat lost by the hotter bodies equals the heat gained by the colder bodies. It is exactly true only if no heat is exchanged with the surroundings.
$2.0(70 - T) = 3.0(T - 20) \Rightarrow 200 = 5.0\,T \Rightarrow T = 40\ ^\circ\text{C}$ .
$m \times 4200 \times 35 = 1.2 \times 4200 \times 40 \Rightarrow m = \frac{48}{35} \approx 1.4\ \text{kg}$ .
$0.50 \times 900 \times (100 - T) = 0.40 \times 4200 \times (T - 20) \Rightarrow 450(100 - T) = 1680(T - 20) \Rightarrow T = \frac{78\,600}{2130} \approx 36.9\ ^\circ\text{C}$ .
Gained: $(0.20 \times 4200 + 0.050 \times 900) \times 3.5 = 885 \times 3.5 = 3097.5\ \text{J}$ . Lost: $0.10 \times c \times 70.5 = 7.05\,c$ . So $c \approx 440\ \mathrm{J\,kg^{-1}\,K^{-1}}$ : probably iron (450).
$0.80 \times 450 \times (400 - T) = 5.0 \times 4200 \times (T - 20) \Rightarrow 360(400 - T) = 21\,000(T - 20) \Rightarrow T = \frac{564\,000}{21\,360} \approx 26.4\ ^\circ\text{C}$ .
(a) The metal's real starting temperature is lower than the recorded one, so the temperature drop used in the calculation is too big; dividing the heat gained by too big a drop gives too small a $c$ . (b) Heat escapes, so the water's rise and the heat gained are too small, which again gives too small a $c$ .

Key points

With no heat lost to the surroundings, heat lost by hot bodies = heat gained by cold bodies.
Write each term as $mc \times$ (higher temperature − lower temperature), and include the calorimeter.
The method of mixtures finds an unknown $c$ ; errors from heat loss usually make it too low.
The final temperature always lies between the starting temperatures.

Sources and credits

14.2: Temperature Change and Heat Capacity (calorimetry), from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Consulted for the physics; no text was copied.
Photo: A Blacksmith in Rwanda.jpg by Germain92, Wikimedia Commons, dedicated to the public domain under CC0 1.0.
Video: Specific heat capacity of solid and liquid, Amrita Create, embedded from YouTube.

Written by Ganzaa: the lesson text, the diagram, the table, worked examples and practice exercises.

1Getting the bath water right

2The principle of heat exchange

3Experiment: the specific heat capacity of a metal by the method of mixtures

4Getting a good result

Getting the bath water right

The principle of heat exchange

Experiment: the specific heat capacity of a metal by the method of mixtures

Getting a good result