Thermodynamics

JEE Main level, single correct, medium

20 of 190

Question 20 of 190

Density of water at 4∘C4{ }^{\circ} \mathrm{C} and 20∘C20^{\circ} \mathrm{C} are 1000 kg/m31000 \mathrm{~kg} / \mathrm{m}^{3} and 998 kg/m3998 \mathrm{~kg} / \mathrm{m}^{3} respectively. The increase in internal energy of 4 kg of water when it is heated from 4∘C4{ }^{\circ} \mathrm{C} to 20∘C20^{\circ} \mathrm{C} is ____\_\_\_\_ J.
(specific heat capacity of water =4.2 J/kg=4.2 \mathrm{~J} / \mathrm{kg}. and 1 atmospheric pressure =105 Pa=10^{5} \mathrm{~Pa})

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Answer: (D) 268799.2

Heat absorbed by water: Q=mcΔT=4×4200×16=268800Q = mc\Delta T = 4 \times 4200 \times 16 = 268800 J.
Volume change: V1=4/1000=0.004V_1 = 4/1000 = 0.004 m³ at 4°C, V2=4/998≈0.00401V_2 = 4/998 ≈ 0.00401 m³ at 20°C.
Work done by water at constant pressure: W=PatmΔV=105×0.00001=0.8W = P_{atm}\Delta V = 10^5 \times 0.00001 = 0.8 J.
By first law of thermodynamics: ΔU=Q−W=268800−0.8=268799.2\Delta U = Q - W = 268800 - 0.8 = 268799.2 J

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