A container with 1kg 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 700Wm−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 ∘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 =0.001s−1, Heat capacity of water =4200J kg−1K−1)
dQdt=eσA(T4−T40)
For small temperature change.
dQdt=eσAT3ΔT ___(i)
mcdTdt=eσAT3ΔT
dTdt=eσAT3mCΔT
eσAT3mC→ constant for Newton law of cooling
eσAT3mC=0.001
eσAT3=mC×0.001
=1×4200×0.001
eσAT3=4.2 ___(ii)
dQdt=700×0.05=35Watts ___(iii)
Put the value of equation (ii), and equation (iii)
in equation (i)
35=4.2ΔT
354.2=ΔT
ΔT=8.33