Question

JEE ADVANCED PHYSICS QUESTION SEPTEMBER 2020

Answer 1

Answer:

A container with '=4200Jkg^(-1)K^(-1)' 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 '=0.001s^(-1),' and it is absorbed by the water over an effective area of '^(@)C' Assuming that the heat loss from the water to the surroundings is governed by Newton's law of cooling, the difference (in '0.05m^(2).' ) 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 '700Wm^(-2)' Heat capacity of water '1kg' )

Answer 2

A container with 1 kg of water in it is kept in sunlight , which cause the water to get warmer than surroundings. the average energy per time per area received due to the sunlight is 700 W/m² and it is absorbed by the water over an effective area is 0.05 m². assuming that the heat loss from the water to the surrounding is governed by Newton's law of cooling , the difference in temperature of water and the surroundings after a long time will be ....

solution : we know, rate of heat , dQ/dt = σAe(T⁴ - T₀⁴) .....(1)

taking T = T₀ + ∆T

so dQ/dt = σAe [(T₀ + ∆T)⁴ - T₀⁴]

= σAeT₀⁴[(1 + ∆T/T₀)⁴ - 1]

from binomial expansion, (1 + ∆T/T₀)⁴ ≈1 + 4∆T/T₀

= 4σAeT₀³∆T .......(2)

but we know, Q = ms∆T

or, dQ/dt = ms dT/dt

⇒ms dT/dt = 4σAeT₀³ ∆T

⇒dT/dt = (4σAeT₀³/ms)∆T

let (4σAeT₀³/ms) = K

here K is constant of Newton's law of cooling

so, K = (4σAeT₀³/ms) = 0.001 /s

⇒4σAeT₀³ = 0.001 × ms = 0.001 × 1 × 4200 = 4.2

here dQ/dt = 700 × 0.05 = 35 Watt

now, 35 = 4.2 × ∆T [ from equation (2) ]

∆T = 35/4.2 = 350/42 = 8.33

Therefore the difference in temperature of water and surroundings is 8.33