3.6.2.1
Continuous flow calorimetry eliminates heat losses
Thermal Energy Transfer — AQA A-Level Physics
- A fluid flows continuously over a heating element.
- The temperature rise (T_2 - T_1) is measured using thermometers at inlet and outlet.
- Two experiments are run with different flow rates but the same temperature rise.
- Heat lost to surroundings is assumed constant for both runs, so it cancels when the equations are subtracted.
Derivation
- Run 1: electrical $energyThe capacity to do work. Measured in joules (J). = energyThe capacity to do work. Measured in joules (J).$ to heat fluid + energyThe capacity to do work. Measured in joules (J). lost.
$$I_1 V_1 t_1 = m_1 c \Delta\theta + E_{\text{lost}}$$
- Run 2: different flow rate and powerThe rate of energy transfer. Measured in watts (W)., same delta theta.
$$I_2 V_2 t_2 = m_2 c \Delta\theta + E_{\text{lost}}$$
- Subtract run 1 from run 2. E_lost cancels:
$$c = \frac{Q_2 - Q_1}{(m_2 - m_1)\Delta\theta}$$
- $c$: specific heat capacityThe energy required to raise the temperature of 1 kg of a substance by 1 K (or 1 °C). of the fluid (J kg^{-1} \(K^{-1}\))
- $Q_1$: electrical energy in run 1 (J)
- $Q_2$: electrical energy in run 2 (J)
- $m_1$: mass of fluid in run 1 (kg)
- $m_2$: mass of fluid in run 2 (kg)
- $\Delta\theta$: temperature rise, same in both runs (K)
Worked Example
Two experiments with t = 40 s each. Run 1: V = 14.0 V, I = 3.0 A, m = 136.0 g. Run 2: V = 9.0 V, I = 3.0 A, m = 73.0 g. Temperature rise = 11 degrees C in both. Calculate c.
Show Solution
1
Calculate Q for each run
- $Q_1 = I_1 V_1 t_1 = 3.0 \times 14.0 \times 40 = 1680$ J
- $Q_2 = I_2 V_2 t_2 = 3.0 \times 9.0 \times 40 = 1080$ J
2
Apply the equation
$$c = \frac{Q_2 - Q_1}{(m_2 - m_1)\Delta\theta} = \frac{1080 - 1680}{(0.073 - 0.136) \times 11}$$
3
Calculate
$$c = \frac{-600}{-0.063 \times 11} = \frac{-600}{-0.693} = 866 \text{ J kg}^{-1} \text{ degrees C}^{-1}$$
Answer
$c = 866$ J kg$^{-1}$ degrees C$^{-1}$