Q: 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^{2} and it is absorbed by the water over an effective area of 0.05 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} , Heat capacity of water = 4200 JKg^{-1} K^{-1} )

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Sol: $\displaystyle \frac{dQ}{dt} = e \sigma A(T^4 – T_o^2)$

For small temperature change ,

$\displaystyle \frac{dQ}{dt} = e \sigma A T^3 \Delta T $ …(i)

$\displaystyle \frac{mC dT}{dt} = e \sigma A T^3 \Delta T $

$\displaystyle \frac{dT}{dt} = \frac{e \sigma A T^3}{m C} \Delta T $

$\displaystyle \frac{e \sigma A T^3}{m C} $ -> Constant for Newton’s Law of cooling

$\displaystyle \frac{e \sigma A T^3}{m C} = 0.001 $

$\displaystyle e \sigma A T^3 = m c \times 0.001 = 1 \times 4200 \times 0.001 $

$\displaystyle e \sigma A T^3 = 4.2 $ …(ii)

$\displaystyle \frac{dQ}{dt} = 700 \times 0.05 = 35 W $ ….(iii)

Putting the value of Eqs (ii) & (iii) in Eq.(i)

35 = 4.2 ΔT

ΔT = 35/4.2 = 8.33