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← JEE Advanced 2020 · Paper 2
PhysicsNumerical+4Question 18 of 54

JEE Advanced 2020 · Paper 2, Physics question 18

A container with 1 kg of water in it is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received due to the sunlight is 700 W m−2700\ \mathrm{W\,m^{-2}} and it is absorbed by the water over an effective area of 0.05 m20.05\ \mathrm{m^2}. Assuming that the heat loss from the water to the surroundings is governed by Newton's law of cooling, the difference (in °C) in the temperature of water and the surroundings after a long time will be ______. (Ignore effect of the container, and take constant for Newton's law of cooling =0.001 s−1= 0.001\ \mathrm{s^{-1}}, Heat capacity of water =4200 J kg−1 K−1= 4200\ \mathrm{J\,kg^{-1}\,K^{-1}})

Answer and worked solution

Official answer: 8.32

After a long time the water's temperature is steady, so the power absorbed equals the heat lost.

Absorbed. 700×0.05=35700\times0.05=35 W.

Lost (Newton's law of cooling):
dQdt=kmc ΔT=0.001×1×4200 ΔT=4.2 ΔT.\frac{dQ}{dt}=kmc\,\Delta T=0.001\times1\times4200\,\Delta T=4.2\,\Delta T.

ΔT=354.2=8.33 ∘C.\Delta T=\frac{35}{4.2}=8.33\ ^\circ\text{C}.
Answer: 8.33 (key: 8.32 to 8.34)