Worked examples · Sheet 1

Equilibrium temperatures in a calorimeter

Theory used: Heat Heat capacity and specific heat Equilibrium temperature Real calorimeter and water equivalent

Exercise 1

$0.30 kg$ of water at $80^{\circ} C$ and $0.50 kg$ of water at $20^{\circ} C$ are poured into an ideal calorimeter. Compute the equilibrium temperature.

Q_{1} + Q_{2} = 0

m_{1} c_{a} (T_{e} - T_{1}) + m_{2} c_{a} (T_{e} - T_{2}) = 0

T_{e} = \frac{m _{1} T _{1} + m _{2} T _{2}}{m _{1} + m _{2}}

T_{e} = \frac{0.30 \cdot 80 + 0.50 \cdot 20}{0.30 + 0.50}^{\circ} C = \frac{34}{0.80}^{\circ} C = 42.5^{\circ} C

The calorimeter is ideal, that is, it has adiabatic walls and negligible heat capacity, therefore the heat released by the hot water is entirely absorbed by the cold water and the sum of the heats exchanged is zero. Each heat is written with the law of calorimetry $Q = m c Δ T$ , with $Δ T = T_{e} - T_{initial}$ ; in this way the sign is automatic, because $Q_{1} < 0$ for the water that cools down and $Q_{2} > 0$ for the water that warms up. The two masses are of the same substance, therefore the specific heat $c_{a}$ cancels and the equilibrium temperature is the average of the temperatures weighted by the masses. The temperatures appear only in differences, therefore they can be left in degrees Celsius.

As a check, $42.5^{\circ} C$ lies between $20^{\circ} C$ and $80^{\circ} C$ and is closer to $20^{\circ} C$ , the temperature of the larger mass.

Exercise 2

A copper block of mass $0.200 kg$ at a temperature of $150^{\circ} C$ is immersed in $0.400 kg$ of water at $20.0^{\circ} C$ , contained in an aluminum vessel of mass $0.100 kg$ that is at the temperature of the water. The vessel is thermally insulated from the outside. Compute the equilibrium temperature and the heat released by the copper. Specific heats: $c_{Cu} = 385 J/ (kg \cdot K)$ , $c_{Al} = 900 J/ (kg \cdot K)$ , $c_{a} = 4186 J/ (kg \cdot K)$ .

m_{Cu} c_{Cu} (T_{e} - T_{Cu}) + (m_{a} c_{a} + m_{Al} c_{Al}) (T_{e} - T_{0}) = 0

T_{e} = \frac{m _{Cu} c _{Cu} T _{Cu} + ( m _{a} c _{a} + m _{Al} c _{Al} ) T _{0}}{m _{Cu} c _{Cu} + m _{a} c _{a} + m _{Al} c _{Al}}

m_{Cu} c_{Cu} = 77.0 J/K, m_{a} c_{a} = 1674.4 J/K, m_{Al} c_{Al} = 90.0 J/K

T_{e} = \frac{77.0 \cdot 150 + ( 1674.4 + 90.0 ) \cdot 20.0}{77.0 + 1674.4 + 90.0}^{\circ} C = \frac{46 838}{1841.4}^{\circ} C \approx 25.4^{\circ} C

Q_{Cu} = m_{Cu} c_{Cu} (T_{e} - T_{Cu}) = 77.0 \cdot (25.4 - 150) J \approx - 9.59 \cdot 1 0^{3} J

The system formed by copper, water and vessel is isolated, therefore the sum of the heats exchanged by the three bodies is zero. The water and the vessel start from the same temperature $T_{0} = 20.0^{\circ} C$ and reach the same $T_{e}$ , therefore their terms are grouped with the total heat capacity $m_{a} c_{a} + m_{Al} c_{Al}$ . Solving the linear equation in $T_{e}$ gives the equilibrium temperature, which is the average of the initial temperatures weighted by the heat capacities.

The heat of the copper is computed with the law of calorimetry; the negative sign indicates that the copper releases about $9.6 kJ$ . The heat capacity of the copper is about twenty-three times smaller than that of the water and vessel, therefore the equilibrium temperature stays close to $20^{\circ} C$ , as the result confirms.

Exercise 3

A calorimeter contains $0.200 kg$ of water at $20.0^{\circ} C$ . $0.100 kg$ of water at $60.0^{\circ} C$ is added and the equilibrium temperature turns out to be $32.0^{\circ} C$ .

Determine the water equivalent of the calorimeter.
The calorimeter, emptied and brought back to $20.0^{\circ} C$ , now contains $0.250 kg$ of water at $20.0^{\circ} C$ . A metal block of mass $0.150 kg$ at a temperature of $100.0^{\circ} C$ is immersed in it, and the equilibrium temperature is $24.0^{\circ} C$ . Determine the specific heat of the metal.

(m_{1} + m_{e}) c_{a} (T_{e} - T_{1}) + m_{2} c_{a} (T_{e} - T_{2}) = 0

m_{e} = \frac{m _{2} ( T _{2} - T _{e} )}{T _{e} - T _{1}} - m_{1}

m_{e} = (\frac{0.100 \cdot 28.0}{12.0} - 0.200) kg = (0.2333 - 0.200) kg \approx 0.0333 kg

(m_{a} + m_{e}) c_{a} (T_{e}^{'} - T_{1}) + m_{x} c_{x} (T_{e}^{'} - T_{x}) = 0

c_{x} = \frac{( m _{a} + m _{e} ) c _{a} ( T _{e}^{'} - T _{1} )}{m _{x} ( T _{x} - T _{e}^{'} )}

c_{x} = \frac{0.2833 \cdot 4186 \cdot 4.0}{0.150 \cdot 76.0} J/ (kg \cdot K) = \frac{4744}{11.4} J/ (kg \cdot K) \approx 416 J/ (kg \cdot K)

The real calorimeter absorbs heat, and this is accounted for with the water equivalent $m_{e}$ ; in the balance the calorimeter behaves like a mass $m_{e}$ of water at the initial temperature of the water it contains, that is, $T_{1} = 20.0^{\circ} C$ . In the first experiment all the substances are water, therefore $c_{a}$ cancels and the only unknown is $m_{e}$ . The result, $33 g$ , says that the vessel with its accessories absorbs heat like $33 g$ of water.

In the second experiment the calorimeter and the water both start from $20.0^{\circ} C$ , therefore they appear together with the mass $m_{a} + m_{e}$ . The balance equation is linear in the unknown $c_{x}$ , which is isolated by dividing by $m_{x} (T_{x} - T_{e}^{'})$ . The metal has specific heat $c_{x} \approx 416 J/ (kg \cdot K)$ , a value between that of copper and that of iron and therefore plausible for a metal. Neglecting the water equivalent would have given $367 J/ (kg \cdot K)$ , with an error of about twelve percent.