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

`t=(21xx10^5)/800=2625s`

`rArrt=2625/60=43.75minapprox44min`