Quantity of Heat and Heat Transfer

Lessons

Senior 3 · Physics

Quantity of Heat and Heat Transfer

5 lessons

Lesson 1 of 5

Heat Capacity, Specific Heat Capacity and Quantity of Heat

25 min

Hot sand, cool water

On a sunny afternoon at the lakeshore in Rubavu, the sand is too hot to walk on barefoot, yet the water of Lake Kivu a few metres away feels cool. Both have been in the same sunshine all day. After sunset the order is reversed: the sand cools quickly, while the lake stays warm long into the night.

In S1 you learned that heat is energy that flows from a hotter body to a colder one, and that temperature tells us how hot a body is. This lesson asks a sharper question: how much heat does it take to warm a given body by a given number of degrees, and why is the answer so different for sand and for water?

What decides the quantity of heat?

Anyone who cooks on a charcoal stove already knows three facts, even without the physics words:

Mass. A full saucepan of water takes longer to heat than half a saucepan on the same stove. The heat needed is proportional to the mass, $Q \propto m$ .
Temperature rise. Warming water from 20 °C to 80 °C takes about twice as long as warming it from 20 °C to 50 °C. The heat needed is proportional to the temperature rise, $Q \propto \Delta T$ .
Material. The same mass of cooking oil gets hot about twice as fast as water. The heat needed depends on what the body is made of.

Putting the three together gives the equation for the quantity of heat $Q$ that a body gains or loses when its temperature changes by $\Delta T$ :

Q = m\,c\,\Delta T \qquad \text{where } \Delta T = T_{\text{final}} - T_{\text{initial}}

The constant $c$ belongs to the material; it is called the specific heat capacity. A change of 1 °C is the same size as a change of 1 K, so $\Delta T$ has the same value whether you measure temperatures in degrees Celsius or in kelvin.

Definition

Heat capacity ( $C$ ) of a body is the quantity of heat needed to raise its temperature by 1 K (or 1 °C): $C = \frac{Q}{\Delta T}$ . SI unit: the joule per kelvin ( $\mathrm{J\,K^{-1}}$ ). It belongs to a particular object. Example: a 1.5 kg iron cooking pot has $C = 1.5 \times 450 = 675\ \mathrm{J\,K^{-1}}$ .

Specific heat capacity ( $c$ ) of a substance is the quantity of heat needed to raise the temperature of 1 kg of it by 1 K: $c = \frac{Q}{m\,\Delta T}$ . SI unit: the joule per kilogram per kelvin ( $\mathrm{J\,kg^{-1}\,K^{-1}}$ ). It belongs to a material. Example: for water $c = 4200\ \mathrm{J\,kg^{-1}\,K^{-1}}$ , so 4200 J warms 1 kg of water by 1 K.

The two are linked by $C = m\,c$ .

Specific heat capacities of some common substances (approximate values)

Substance	$c$ ( $\mathrm{J\,kg^{-1}\,K^{-1}}$ )	In practice
Water	4200	Slow to warm and to cool; stores and carries heat well
Ice	2100	Half the value for liquid water
Cooking oil	about 2000	Heats about twice as fast as the same mass of water
Aluminium	900	Light saucepans heat quickly
Glass	about 840	Used for beakers and bottles
Dry sand	about 800	Heats up and cools down quickly
Iron	450	Cooking pots, stoves
Copper	390	Heat-exchanger pipes, the bottoms of good pans
Lead	130	Among the lowest of common metals

Water has the highest value in the table by far. That single fact explains the beach in Rubavu: in the same sunshine, each kilogram of sand needs only about a fifth as much heat as a kilogram of water to warm by one degree.

Worked example 1: heating water for tea

How much heat is needed to warm 2.5 kg of water from 20 °C to 90 °C?

\Delta T = 90 - 20 = 70\ \text{K}

Q = m\,c\,\Delta T = 2.5 \times 4200 \times 70 = 735\,000\ \text{J} = 735\ \text{kJ}

Worked example 2: how long does the kettle take?

A 2000 W electric kettle holds 1.2 kg of water at 20 °C. Assuming no heat is lost, how long does it take to heat the water to 95 °C?

Q = 1.2 \times 4200 \times (95 - 20) = 378\,000\ \text{J}

Power is energy per second, so

t = \frac{Q}{P} = \frac{378\,000}{2000} = 189\ \text{s} \approx 3.2\ \text{min}

A real kettle takes a little longer, because some heat also warms the kettle itself and some escapes to the air.

Experiment: measuring the specific heat capacity of aluminium

You need: a 1.0 kg aluminium block with two holes, a 12 V immersion heater, a 12 V d.c. supply, an ammeter, a voltmeter, a thermometer, lagging (felt or foam), a stop-clock and a few drops of cooking oil.

Wrap the block in lagging. Put the heater in one hole and the thermometer in the other, with a drop of oil in the thermometer's hole so it touches the metal well.
Read the starting temperature $T_{1}$ .
Switch on and start the clock. Record the voltmeter reading $V$ and ammeter reading $I$ ; keep them steady.
After a measured time $t$ , switch off. Keep watching the thermometer and record the highest temperature $T_{2}$ it reaches (the heater is still hotter than the block for a short time).

The electrical energy supplied is $E = V\,I\,t$ . If none of it escapes, all of it warms the block, so

V\,I\,t = m\,c\,(T_2 - T_1) \quad\Rightarrow\quad c = \frac{V\,I\,t}{m\,(T_2 - T_1)}

Important

Safety. Use only the low-voltage (12 V) supply, never a mains heater. Never switch the heater on when it is out of the block: it overheats in seconds. After the experiment the heater stays hot for several minutes; lift it by its leads, not its metal case.

Worked example 3: the experiment's results

In the experiment above, $V = 12.0\ \text{V}$ , $I = 4.0\ \text{A}$ , $t = 5.0\ \text{min}$ , $m = 1.0\ \text{kg}$ , and the temperature rose from 22.0 °C to 38.0 °C. Find $c$ for aluminium.

E = V\,I\,t = 12.0 \times 4.0 \times 300 = 14\,400\ \text{J}

c = \frac{14\,400}{1.0 \times (38.0 - 22.0)} = \frac{14\,400}{16} = 900\ \mathrm{J\,kg^{-1}\,K^{-1}}

Without lagging, some of the 14 400 J would escape to the air, the block would warm less, and the calculated $c$ would come out too large.

Warning

Three common mistakes. (1) Using grams: in $Q = mc\,\Delta T$ with $c$ in $\mathrm{J\,kg^{-1}\,K^{-1}}$ , the mass must be in kilograms, so 250 g is 0.25 kg. (2) Using the final temperature instead of the change in temperature. (3) Adding 273 to a temperature change: a rise of 30 °C is a rise of 30 K, not 303 K.

Why water's high specific heat capacity matters

Climate by the lakes. Give 10 kJ to 1 kg of dry sand and it warms by $\frac{10\,000}{800} = 12.5\ \text{K}$ ; give the same to 1 kg of water and it warms by only $\frac{10\,000}{4200} \approx 2.4\ \text{K}$ . Large lakes such as Kivu therefore warm and cool slowly, which keeps the temperature of the land beside them more even.
Cooling engines. Water pumped round a car or generator engine carries away a lot of heat for a small rise in its own temperature, then gives it up to the air in the radiator.
Storing heat. A hot-water bottle or a flask of hot water stays warm for a long time because each kilogram holds so much heat.
Your body. About 60% of an adult's body mass is water, which helps to keep body temperature steady.

A low specific heat capacity is useful too: a copper or aluminium pan warms quickly and wastes little of the stove's heat on itself.

Tip

Small c, capital C. Small $c$ is for a material (all water, anywhere): per kilogram. Capital $C$ is for a particular object (this kettle): the whole thing. Multiply: $C = m\,c$ .

Specific heat capacity explained, measured and demonstrated (fizzics.org). Watch how the energy supplied to a known mass of water and its temperature rise give the specific heat capacity of water: the same method as the aluminium block above.

✦

Practice

Define specific heat capacity and give its SI unit. How is it different from heat capacity?
How much heat is needed to warm 0.50 kg of water from 20 °C to 80 °C?
An aluminium saucepan has a mass of 2.0 kg. Find (a) its heat capacity; (b) the heat needed to raise its temperature by 50 °C.
A 0.40 kg copper block absorbs 6240 J of heat. By how much does its temperature rise?
A 50 W heater warms 0.80 kg of cooking oil from 24.0 °C to 31.5 °C in 4.0 min. Assuming no heat is lost, find the specific heat capacity of the oil.
A 1500 W kettle is made of 0.60 kg of aluminium and holds 1.5 kg of water. Both start at 20 °C. How long does it take to bring kettle and water to 80 °C if no heat is lost?
Explain why, on a sunny day, the sand of a beach becomes much hotter than the water of the lake beside it.
In the aluminium-block experiment, a pupil forgets the lagging. Will her value of $c$ be too high or too low? Explain.

Show answersHide answers

The heat needed to raise the temperature of 1 kg of a substance by 1 K; $\mathrm{J\,kg^{-1}\,K^{-1}}$ . Heat capacity is for a whole object (heat per kelvin, $\mathrm{J\,K^{-1}}$ ), and $C = m c$ .
$Q = 0.50 \times 4200 \times 60 = 126\,000\ \text{J} = 126\ \text{kJ}$ .
(a) $C = 2.0 \times 900 = 1800\ \mathrm{J\,K^{-1}}$ ; (b) $Q = 1800 \times 50 = 90\,000\ \text{J} = 90\ \text{kJ}$ .
$\Delta T = \frac{6240}{0.40 \times 390} = 40\ \text{K}$ (40 °C).
$Q = 50 \times 240 = 12\,000\ \text{J}$ ; $c = \frac{12\,000}{0.80 \times 7.5} = 2000\ \mathrm{J\,kg^{-1}\,K^{-1}}$ .
$Q = (1.5 \times 4200 + 0.60 \times 900) \times 60 = (6300 + 540) \times 60 = 410\,400\ \text{J}$ ; $t = \frac{410\,400}{1500} \approx 274\ \text{s} \approx 4.6\ \text{min}$ .
Sand has a much lower specific heat capacity (about 800) than water (4200), so the same heat per kilogram raises its temperature about five times as much. The water is also stirred by waves and currents, so its heat is shared through a large mass.
Too high. Heat escapes to the air, so the block's temperature rise is smaller than it should be for the energy $V I t$ supplied. Dividing $V I t$ by a smaller $m\,\Delta T$ gives a larger $c$ .

Key points

Quantity of heat: $Q = m\,c\,\Delta T$ , with $m$ in kg and $\Delta T$ in K (or °C).
Specific heat capacity $c$ belongs to a material ( $\mathrm{J\,kg^{-1}\,K^{-1}}$ ); heat capacity $C = m c$ belongs to an object ( $\mathrm{J\,K^{-1}}$ ).
Electrical method: $c = \frac{V\,I\,t}{m\,\Delta T}$ ; lagging reduces heat loss, which would make $c$ too large.
Water's high $c$ (4200) makes it slow to warm, slow to cool and ideal for carrying heat.

Sources and credits

14.2: Temperature Change and Heat Capacity, 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.
Specific heat capacities are rounded values from standard physical data tables.
Photo: Gisenyi (6817417653).jpg by Martijn Munneke, Wikimedia Commons, licensed under CC BY 2.0.
Video: Specific heat capacity explained, measured and demonstrated, fizzics.org, embedded from YouTube.

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

1Hot sand, cool water

2What decides the quantity of heat?

3Specific heat capacities of some common substances (approximate values)

4Experiment: measuring the specific heat capacity of aluminium

5Why water's high specific heat capacity matters

Hot sand, cool water

What decides the quantity of heat?

Specific heat capacities of some common substances (approximate values)

Experiment: measuring the specific heat capacity of aluminium

Why water's high specific heat capacity matters