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Question

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℃) in the temperature of water and the surroundings after a long time will be _____________.

(Ignore the effect of the container, and take constant for Newton’s law of cooling =0.001s-1, the Heat capacity of water =4200Jkg-1K-1)


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Solution

Step 1. Given data:

m=1kg Weight of water.

A=0.05m2, Effective area of water

C=4200JKg-1K-1, Heat capacity of water

1AdQdt=700Wm-2, Energy per unit time per unit area

⇒dQdt=700×0.05∴dQdt=35W

eσAT3mC=0.001s-1, Constant for Newton's law of cooling

eσAT3=0.001×1×4200eσAT3=4.2

Step 2. Calculating value of the difference (in℃) in the temperature of water and the surroundings after a long time

dQdt=eσAT4-T4o

For small temperature change,

dQdt=eσAT3∆T

Putting all values in the above equation,

dQdt=eσAT3∆T⇒35=4.2×∆T⇒∆T=354.2∴∆T=8.33

Hence, the difference (in℃) in the temperature of water and the surroundings after a long time ∆T=8.33.


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