Physics · Molecular physics and thermodynamics · Grade 7

Calorimeter: heat balance

If we add cold water to hot tea, what will the temperature be, and where does the heat go?

The situation

The tea in your cup is very hot — you add a little cold water to it. What will the temperature of the mixture be? The cold water takes the heat that the tea gives up, but part of it warms the cup itself and part escapes into the air. A calorimeter is a vessel that “counts” heat: in it we measure the temperature of mixtures, the specific heat capacity of metals, the heat given out by fuel, and even how much heat a given amount of mechanical work turns into. How much 20 °C water must be added to tea at 80 °C to bring it down to 40 °C?

Topic

Quantity of heat and the heat balance. Mixing tea and cold water: specific heat capacity in a calorimeter

Key ideas

Quantity of heat
Heat balance equation
Specific heat capacity

Formulas

Q = c m (T_{2} - T_{1})

Quantity of heat. For water $c = 4186$ J/(kg·K) (rounded to 4200 in textbooks).

c m_{1} (T_{1} - θ) = (c m_{2} + C_{k}) (θ - T_{2})

Heat balance: heat given up by the tea = heat received by the cold water and the calorimeter vessel. $C_{k} = c_{A l} m_{c u p} = 90$ J/K.

c_{x} = \frac{(c m_{s} + C_{k}) (θ - T_{s})}{m_{x} (T_{x} - θ)}

Lab work: specific heat capacity of a sample heated to 100 °C.

Q = q m_{y o}, η = \frac{c m Δ T}{q m_{y o}}

Heat released when fuel burns, and the efficiency of a heater.

Δ U = A + Q, A = 2 M g h

The first law of thermodynamics and Joule's experiment: both work and heat change internal energy; 1 cal ≈ 4.19 J.

Theory

Internal energy and quantity of heat

The molecules of a body move randomly and interact with each other: the sum of their kinetic and potential energies is the internal energy $U$ . When hot and cold bodies touch, energy passes from the hot one to the cold one; this transferred energy is the quantity of heat $Q$ (in joules). To heat a body by $Δ T$ requires $Q = c m Δ T$ : $c$ is the specific heat capacity (the heat that warms 1 kg by 1 K), and $C = c m$ is the heat capacity of the body, in J/K. On the graph the line of the hot body goes down, that of the cold body goes up, and they meet at one value $θ$ .

Heat balance equation

In a system isolated from its surroundings energy is not lost: the heat given up by the hot bodies equals the heat received by the cold bodies, $Q_{g i v e n} = Q_{r e c e i v e d}$ (or $\sum Q_{i} = 0$ ). In a calorimeter the inner cup is separated from the outer vessel by a layer of air and an insulating ring, so the loss is small. But the cup itself also warms up: 100 g of aluminium has $C_{k} = 90$ J/K, that is, it is “equivalent” to ≈ 21 g of water. If you forget it, you will predict $θ$ too high. When loss is switched on, after equilibrium the temperature slowly falls; the “loss” on the HUD $= Q_{g i v e n} - Q_{r e c e i v e d}$ .

Laboratory: specific heat capacity of a solid

The sample is heated to 100 °C in boiling water and quickly transferred to the calorimeter. After measuring the equilibrium temperature $θ$ , $c_{x}$ is found from the balance equation and compared with the table: aluminium 900, iron 460, copper 385, tin 230, lead 130 J/(kg·K). Even when the samples have the same mass, their sizes differ — their densities are different. Sources of error: the sample cools during transfer, some boiling water clings to it, and the thermometer has limited accuracy. In the model the sample does not cool while being transferred.

Specific heat of combustion of fuel

The heat released when 1 kg of fuel burns completely is the specific heat of combustion $q$ : dry firewood $1.0 \cdot 10^{7}$ , coal and alcohol $2.7 \cdot 10^{7}$ , natural gas $4.4 \cdot 10^{7}$ J/kg. Not all the heat released reaches the water: hot gases pass by the pot, and the pot itself and the air are heated. Efficiency $η = Q_{w a t e r} / (q m_{f u e l})$ . The efficiencies of the burners in the model are approximate: spirit lamp 30 %, gas burner 45 %, open wood fire 15 %, coal 25 %; because of losses from the pot to the air, the measured efficiency turns out even smaller.

Joule's experiment: work → heat

James Joule heated water by stirring it with a paddle wheel turned by falling weights. When the weights fall from a height $h$ , the work done is $A = 2 M g h$ ; the paddles “rub” the water and turn this work into internal energy. The first law of thermodynamics: $Δ U = A + Q$ — internal energy can be changed both by doing work and by transferring heat. When a gas expands, it itself does thermodynamic work $A = p Δ V$ . From Joule's experiment comes the mechanical equivalent of heat: 1 cal (the heat that warms 1 g of water by 1 °C) ≈ 4.19 J. The calculation also takes into account the vessel and paddle (46 cal/°C) and the energy $\frac{1}{2} \cdot 2 M v^{2}$ lost when the weights hit the floor.

Heat capacity of liquids and gases

The $c$ of water is very large — 9 times that of iron: part of the energy goes into the hydrogen bonds between molecules. That is why water is a good heat carrier (heating systems, car radiators). When a liquid is heated its volume hardly changes, $c_{p} \approx c_{V}$ . For gases the difference is large: a gas heated at constant pressure expands and does work, so $C_{p} - C_{V} = R$ (Mayer's relation); for a monatomic ideal gas $C_{V} = \frac{3}{2} R$ , for a diatomic one $\frac{5}{2} R$ , and $γ = C_{p} / C_{V}$ determines the adiabatic process ( $p V^{γ} = const$ ). The internal energy of an ideal gas depends only on temperature, $U = \frac{i}{2} ν R T$ : when Joule let a gas expand into a vacuum (no work and no heat) its temperature did not change. Isoprocesses, the adiabat (Poisson's equation) and finding $γ$ by the Clément–Desormes method are in the “Gas under a piston” experiment; the Joule–Thomson process is in the “Real gases” experiment; fluctuations are in a separate experiment.

Did you know?

In the 1840s James Joule measured the mechanical equivalent of heat by heating water with a paddle wheel turned by a falling weight. The unit of energy, the joule, is named after him.
Seas and lakes warm up slowly in summer and cool down slowly in winter: the specific heat capacity of water is several times that of soil and rock. That is why the climate near large bodies of water is milder.

Inside the experiment

The experiment starts from a real-life situation: you predict the result, change the controls, run the experiment and draw a conclusion. Inside are missions, theory, a quiz and a guided tour in which the camera moves and explains the key points.

More experiments in this section

Melting, boiling and evaporation — Grade 7
Tandoor and sandal: heat — Grade 7
Gas under a piston — Grade 9
Humidity and dew — Grade 9
Heat engine: the Otto cycle — Grade 9
Entropy and the arrow of time — Grade 9
Brownian motion and diffusion — Grade 9
Surface tension and capillarity — Grade 9
Crystalline and amorphous solids — Grade 9
Phase transitions — Grade 10