The Principle of Calorimetry

Energy is conserved in a closed box

Pour hot water into cold water in a steel flask with its lid on. Heat flows from the hot water to the cold until both reach one temperature, and then it stops. The chapter on temperature called this state thermal equilibrium. If no heat leaks out of the flask, energy is conserved inside it. Every joule the hot water loses, the cold water gains.

The principle is energy conservation, written for heat. It needs two conditions. First, the vessel must be insulated, so that no heat is exchanged with the room. Second, nothing inside may make or swallow heat unseen, such as a chemical reaction or a change of state left out of the account. A change of state is allowed, but its \(mL\) must be counted, as the next lesson does.

The final temperature of a mixture

Mix a mass \(m_1\) at \(T_1\) with a mass \(m_2\) of the same liquid at a lower \(T_2\). Let the final temperature be \(T\). Heat lost by the hot part is \(m_1 c (T_1 - T)\). Heat gained by the cold part is \(m_2 c (T - T_2)\). Setting them equal gives \(T\) as a weighted mean of the two temperatures.

So 100 g of water at 80 °C mixed with 100 g at 20 °C gives 50 °C, the plain mean. Equal heat capacities always meet at the mean. Unequal ones meet nearer the temperature of the body with the larger heat capacity. Mix 200 g of water at 90 °C with 100 g at 30 °C: \(T = (200 \times 90 + 100 \times 30)/300 = 70\,\text{°C}\).

Work out the figure above. The water has \(C = 0.2 \times 4200 = 840\,\text{J/K}\). The copper block has \(C = 0.2 \times 400 = 80\,\text{J/K}\). So \(920\,T = 80 \times 100 + 840 \times 20\), which gives \(T = 27.0\,\text{°C}\). The aluminium block has \(C = 180\,\text{J/K}\). Then \(1020\,T = 180 \times 100 + 840 \times 20\), and \(T = 34.1\,\text{°C}\). In both, the block falls through many more degrees than the water rises. The same heat passes out of one and into the other, and the water needs far more heat per degree.

The calorimeter

A calorimeter is built to make the principle true. Its inner vessel is thin copper. Copper conducts well, so the vessel and its water reach one temperature quickly. A thin vessel also has a small heat capacity, so it takes little of the heat itself. The vessel stands on insulating supports inside an outer jacket, with an air gap between them. The jacket, the air gap and a lid keep heat from leaking to the room.

Two more parts matter. A stirrer mixes the water, so that all of it is at one temperature before it is read. Without stirring, the thermometer reads only the layer round its bulb. The lid also stops evaporation from the water, which would carry latent heat away unseen. The heat taken by the copper vessel and the stirrer is counted through their water equivalent.

Finding a specific heat by the method of mixtures

To measure the specific heat of a metal, heat a known mass of it to a known temperature, often in boiling water. Drop it quickly into the calorimeter and stir. Read the highest steady temperature. The heat lost by the metal equals the heat gained by the water and the vessel. That equation has one unknown, the metal's \(c\). This is the method of mixtures.

Problems are often set in calories, with water's specific heat 1 cal/(g °C). Then a mass of water in grams is also its heat capacity in cal/°C, and a water equivalent simply adds to it. Take a calorimeter of water equivalent 10 g with 90 g of water at 20 °C. A 100 g metal block at 110 °C is dropped in and the mixture settles at 30 °C. Heat gained: \(100 \times 1 \times 10 = 1000\,\text{cal}\). Heat lost: \(100 \times s \times 80\), so \(s = 0.125\,\text{cal/(g °C)}\), which is 525 J/(kg K).

Three liquids, mixed in pairs

The principle can find a result that was never measured. Suppose equal masses of three liquids A, B and C start at 12 °C, 19 °C and 28 °C. A mixed with B settles at 16 °C. B mixed with C settles at 23 °C. What would A mixed with C give? Each mixture gives one ratio of specific heats, and two ratios fix the third mixture.