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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.
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Solution
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`
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