Verified Solution • 2025–26 Session moderate

QJEEAdv20Ph097e: JEE Advanced 2020 Physics - Class 11 JEEAdv20Ph097e Solution: A container with 1 kg of water in it is kept in sunlight, which …

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 Wm −- 2 and it is absorbed by the water over an effective area of 0.05m2. Assuming that the heat loss from the water to the surroundings is governed by Newton's law of cooling, the difference (in ∘^\circ 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, Heat capacity of water = 4200 J kg −- 1 K −- 1)

Step-by-Step Verified Solution

dQdt=eσA(T4−T04){{dQ} \over {dt}} = e\sigma A({T^4} - T_0^4)

For small temperature change,

dQdt=eσAT3ΔT{{dQ} \over {dt}} = e\sigma A{T^3}\Delta T .... (i)

mCdTdt=eσAT3ΔT{{mCdT} \over {dt}} = e\sigma A{T^3}\Delta T

⇒dTdt=eσAT3mCΔT\Rightarrow {{dT} \over {dt}} = {{e\sigma A{T^3}} \over {mC}}\Delta T

eσAT3mC{{e\sigma A{T^3}} \over {mC}} →\to constant for Newton law of cooling

∴\therefore eσAT3mC=0.001{{e\sigma A{T^3}} \over {mC}} = 0.001

⇒eσAT3=mC×0.001=1×4200×0.001\Rightarrow e\sigma A{T^3} = mC \times 0.001 = 1 \times 4200 \times 0.001

eσAT3=4.2e\sigma A{T^3} = 4.2 .... (ii)

dQdt=700×0.05=35{{dQ} \over {dt}} = 700 \times 0.05 = 35 W ..... (iii)

Putting the value of Eqs. (ii) and (iii) in Eq. (i), we get

35=4.2ΔT⇒354.2=ΔT⇒ΔT=8.3335 = 4.2\Delta T \Rightarrow {{35} \over {4.2}} = \Delta T \Rightarrow \Delta T = 8.33

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