Temperature, heat and specific heat capacity

Temperature and the kinetic particle model, the Kelvin scale, internal energy, heat as energy transferred because of a temperature difference, thermal equilibrium, specific heat capacity and Q = mcΔT, and the practical methods (electrical heating and method of mixtures) for measuring it.

On a summer afternoon at Bondi the sand can be hot enough to burn your feet while the sea is a cool 22 °C, even though both have had the same sunshine all day. Overnight the sand cools right down while the sea barely changes. Sea breezes, Sydney's mild coastal climate compared with Penrith's, and why a car's cooling system uses water all come back to one property of water: it takes a lot of energy to change its temperature.

In Year 8 you used the particle model to explain conduction, convection and radiation. This page makes the model quantitative.

Temperature and the kinetic particle model

The particles of every substance are always moving: vibrating in a solid, sliding past each other in a liquid, flying about in a gas. Temperature is a measure of the average translational kinetic energy of the particles. Heat a gas and its molecules move faster on average; cool it and they slow.

Not every particle has the same energy. At any temperature there is a spread, from a few slow particles to a few very fast ones, and temperature tells you the average. The fast ones explain why a puddle evaporates well below 100 °C.

Lower the temperature and the average kinetic energy falls. The lowest possible temperature, absolute zero, is where the particles have the least possible energy. On the Kelvin scale this is 0 K, which is −273.15 °C. One kelvin is the same size as one degree Celsius, so

$T(\text{K}) = T(°\text{C}) + 273.15$

and a temperature change is the same number in both units: a rise of 15 °C is a rise of 15 K. (Older books say the particles are "completely still" at absolute zero. Quantum physics shows that some motion always remains, so it is better to say they have their minimum energy. Since 2019 the kelvin has been defined by fixing the value of the Boltzmann constant, rather than by the triple point of water.)

Internal energy and heat

The internal energy of an object is the total kinetic and potential energy of all its particles. A bath of warm water has far more internal energy than a cup of boiling water, even though its temperature is lower, because it has so many more particles.

Heat (symbol $Q$, unit J) is energy that is transferred from one object to another because of a temperature difference. It always flows spontaneously from the hotter object to the cooler one. An object does not "contain heat": it contains internal energy, and heat is the name for that energy while it is on the move. Energy can also be transferred by work, as when you rub your hands together or pump up a tyre.

Thermal equilibrium

Put a hot object in contact with a cooler one. Faster particles in the hot object collide with slower ones in the cool object and, on average, pass on energy. The hot object cools and the cool one warms until they reach the same temperature. Then energy still moves both ways at the particle level, but equally, so there is no net transfer. This state is thermal equilibrium.

Two objects in contact approach the same final temperature. The net heat flow stops when their temperatures are equal.

This is what a thermometer relies on: it comes to thermal equilibrium with whatever it measures, so its temperature equals that of the object. If two objects are each in thermal equilibrium with a third, they are in equilibrium with each other (sometimes called the zeroth law of thermodynamics).

Specific heat capacity

The energy needed to change the temperature of an object depends on its mass, the temperature change and the material. The specific heat capacity $c$ of a material is the energy needed to raise the temperature of 1 kg of it by 1 K (or 1 °C).

$Q = mc\Delta T$

with $Q$ in J, $m$ in kg, $\Delta T$ in K or °C, and $c$ in J kg$^{-1}$ K$^{-1}$.

Material	$c$ (J kg$^{-1}$ K$^{-1}$)
Water	4180
Ice	2100
Steam	2010
Aluminium	900
Concrete	about 880
Dry sand	about 800
Iron and steel	450
Copper	385
Lead	130

Water's very high specific heat capacity is why the sea warms and cools so slowly compared with sand, why coastal cities have milder temperatures, and why water is used to carry energy in car radiators and hot-water systems. Sand needs about a fifth of the energy per kilogram to warm by the same amount, so it heats and cools much more quickly.

Worked example: heating water

An electric kettle heats 1.5 kg of water from 18 °C to 100 °C. How much energy is transferred to the water, and how long does it take with a 2400 W element, assuming no losses?

$Q = mc\Delta T = 1.5 \times 4180 \times (100 - 18) = 5.14 \times 10^5$ J.

energy

$t = \dfrac{Q}{P} = \dfrac{5.14 \times 10^5}{2400} = 214$ s, about 3.6 minutes. A real kettle takes a little longer because some energy heats the kettle and the air.

time

Worked example: method of mixtures

A 0.200 kg block of aluminium at 100 °C is dropped into 0.250 kg of water at 20.0 °C in an insulated cup. Find the final temperature. ($c_{\text{Al}} = 900$, $c_{\text{water}} = 4180$ J kg$^{-1}$ K$^{-1}$.)

Energy lost by the aluminium = energy gained by the water (no losses): $m_{\text{Al}}c_{\text{Al}}(100 - T) = m_{\text{w}}c_{\text{w}}(T - 20.0)$.

principle

$0.200 \times 900 \times (100 - T) = 0.250 \times 4180 \times (T - 20.0)$, so $180(100 - T) = 1045(T - 20.0)$.

substitute

$18\,000 - 180T = 1045T - 20\,900$, so $1225T = 38\,900$ and $T = 31.8$ °C.

solve

The final temperature is much closer to the water's starting temperature, because the water's $mc$ (1045 J/K) is almost six times the aluminium's (180 J/K).

check

Measuring specific heat capacity

Electrical method: the energy supplied is Q = VIt.

Investigation: specific heat capacity of a metal (electrical method)

Method: a 1.00 kg metal block has two holes, one for an immersion heater and one for a thermometer (a drop of oil in the hole improves thermal contact). Wrap the block in insulation. Connect the heater to a low-voltage DC supply with an ammeter in series and a voltmeter across the heater. Record the starting temperature, switch on and start a stopwatch. Record $V$, $I$ and the temperature every minute for about 10 minutes.

Processing: energy supplied $Q = VIt$. Plot temperature against time. For the straight part of the graph, $\text{gradient} = \dfrac{\Delta T}{t} = \dfrac{VI}{mc}$, so $c = \dfrac{VI}{m \times \text{gradient}}$.

Typical result: aluminium, 12.0 V and 4.00 A for 300 s gives $Q = VIt = 14\,400$ J. With no losses the block would warm by 16.0 °C; a real run might give 15.5 °C, so $c = \dfrac{14\,400}{1.00 \times 15.5} = 929$ J kg$^{-1}$ K$^{-1}$, a little above the accepted 900.

Errors: energy lost to the surroundings and used to heat the heater and thermometer means the block warms less than it should, so $c$ comes out too high (a systematic error). Insulate the block, use the gradient of the early, straight part of the graph, and keep the temperature rise modest. Also, the temperature keeps rising briefly after switching off, as energy spreads from the heater, so record the maximum.

Safety: the heater and block get hot: handle with tongs and let them cool. Use only low-voltage supplies.

Method of mixtures: a hot sample is transferred to water in an insulated cup.

Method of mixtures: heat a metal sample of known mass in boiling water (so it is at about 100 °C), quickly transfer it into a known mass of water in a polystyrene cup, stir and record the highest temperature. Set energy lost by the metal equal to energy gained by the water, as in the worked example, and solve for $c$. Energy lost in the transfer and from the cup makes the water's rise smaller, so here $c$ comes out too low.

InteractiveCalorimeter lab

Try this

In Method of mixtures with 200 g of aluminium and 250 g of water, press Start: the water settles at 31.8 °C and the calculated c is 900 J/(kg·K), matching the worked example.
Choose copper: the water only reaches 25.5 °C. Explain why using $mc$ for each.
Choose Mystery metal, run it, and identify the metal from the table using your calculated c.
In Electrical heating with aluminium at 4.0 A, run for 300 s: Q = 14 400 J, ΔT = 16.0 °C and c = 900. Turn on Heat loss and repeat. Is c now too high or too low?

Test yourself

How much energy is needed to warm 250 g of copper from 20 °C to 80 °C? ($c = 385$ J kg$^{-1}$ K$^{-1}$)
Two objects are in thermal equilibrium. This means they have the same
Equal masses of water and sand receive the same energy in the sun. The sand's temperature rises about five times as much because
In the electrical method, a student ignores heat lost from an uninsulated block. The calculated specific heat capacity will be

Label the specific heat capacity apparatus.

Which quantity is being described?

Temperature is a measure of the kinetic energy of the particles. Absolute zero is 0 K, or °C. Objects in contact reach thermal when their temperatures are equal. The energy needed to raise 1 kg of a material by 1 K is its specific heat . Water's value is high: about J kg⁻¹ K⁻¹.

6 marks

To find the specific heat capacity of an unknown metal, a student heats a 0.150 kg sample in boiling water to 100 °C and transfers it into 0.200 kg of water at 21.0 °C in a polystyrene cup. The highest temperature reached is 26.5 °C. (a) Calculate the specific heat capacity of the metal. (b) Identify the metal from the table on this page. (c) Analyse the effect of energy losses on the result, and suggest two improvements.

Your answer