Thermal Properties of Matter

JEE Main level, numerical answer, medium

57 of 126

Question 57 of 126

A water heater of power 2000W2000W is used to heat water. The specific heat capacity of water is 4200Jkg−1K−14200J{kg}^{-1}{K}^{-1}. The efficiency of heater is 7070%. Time required to heat 2kg2kg of water from 10∘C10^{\circ}C to 60∘C60^{\circ}C is _____ ss.
(Assume that the specific heat capacity of water remains constant over the temperature range of the water).

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Answer: 300

Heat provided to water will be responsible for increase in temperature. Therefore, we can write
η×P×t=M×s×ΔT\eta \times P\times t=M\times s\times \Delta T
Now,
η=70100=0.7\eta =\dfrac{70}{100}=0.7 and energy=power×time\text{energy}=\text{power}\times \text{time}.
Therefore,

t=2×4200×(60∘−10∘)0.7×2000s=300st=\dfrac{2\times 4200\times \left(60^{\circ}-10^{\circ}\right)}{0.7\times 2000}s=300s

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