Question

Calculate the time required to heat 20 kg of water from 10°C to 35°C using an immersion heater rated 1000 W. Assume that 80% of the power input is used to heat the water. Specific heat capacity of water = 42000 J kg⁻¹ K⁻¹.

Answer 1

Given:
Power rating of the immersion rod, P = 1000 W
Specific heat of water, S = 4200 J kg⁻¹ K⁻¹
Mass of water, M = 20 kg
Change in temperature, ΔT = 25 °C

Total heat required to raise the temperature of 20 kg of water from 10°C to 35°C is given by
Q = M × S × ΔT
Q = 20 × 4200 × 25
Q = 20 × 4200 × 25 = 21 × 10⁵ J

Let the time taken to heat 20 kg of water from 10°C to 35°C be t. Only 80% of the immersion rod's heat is useful for heating water. Thus,
Energy of the immersion rod utilised for heating the water = t × (0.80) × 1000 J

t × (0.80) × 1000 J = 21 × 10⁵ J
$$\begin{gathered} t=\frac{21\times {10}^5}{800}=2625\mathrm{s} \\ \Rightarrow t=\frac{2625}{60}=43.75\min \approx 44\min \end{gathered}$$