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JEE Advanced 2020Paper 2
Exam
JEE Advanced
Year
2020
Paper
Paper 2
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Advanced 2020 · Paper 2 · Physics · Thermal Properties of Matter

A container with 1kg of water in it is kept in sunlight, which...

A container with 1kg1kg 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−2700W{m}^{- 2} and it is absorbed by the water over an effective area of 0.05m2.0 . 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∘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,= 0 . 001{s}^{- 1}, Heat capacity of water =4200Jkg−1K−1= 4200J{kg}^{- 1}{K}^{- 1})

Correct answer

0

Step-by-step solution

Pi=700×0.05=35{P}_{i} = 700 \times 0 . 05 = 35 watts

Pi=Pr{P}_{i} = {P}_{r}

35=e\sigmaA4T03(\DeltaT)...(1)35 = e\sigmaA4{T}_{0}^{3} ( \DeltaT ) . . . \left(1\right)

∴K=4e\sigmaAT03mC=0.0001∴ K = \dfrac{4e\sigmaA{T}_{0}^{3}}{mC} = 0 . 0001

4e\sigmaAT03=0.001×mC=4.24e\sigmaA{T}_{0}^{3} = 0 . 001 \times mC = 4 . 2

from equation (1)

\DeltaT=354.2=506=253=8.33\DeltaT = \dfrac{35}{4 . 2} = \dfrac{50}{6} = \dfrac{25}{3} = 8 . 33

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