Water of mass m gram is slowly heated to increase the temperature from T₁ to T₂. The change in entropy of the water, given specific heat of water is 1 J kg⁻¹ K⁻¹, is: (a) m(T₂ - T₁) (b) zero (c) m ln ((T₁)/(T₂)) (d) m ln ((T₂)/(T₁))

Answer: (d). dS = dQ/T = (mc dT)/(T) Δ S = mc∫T₁T₂ dT/T = mcln (T₂)/(T₁) With c = 1: Δ S = mln (T₂)/(T₁). So the answer is (d) mln ((T₂)/(T₁)).

Water of mass m gram is slowly heated to increase the temperature… | JEE Main 2025 (23 Jan, Shift 2)

Q12JEE Main 2025 · 23 JAN · SHIFT 2Medium

Heat Engines, Refrigerators & Second Law

Water of mass $m$ gram is slowly heated to increase the temperature from $T_1$ to $T_2$ . The change in entropy of the water, given specific heat of water is $1\text{ J kg}^{-1}\text{ K}^{-1}$ , is:

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Answer: (d)

$dS = \frac{dQ}{T} = \frac{mc\,dT}{T}$
$\Delta S = mc\int_{T_1}^{T_2} \frac{dT}{T} = mc\ln\frac{T_2}{T_1}$
With $c = 1$ : $\Delta S = m\ln\frac{T_2}{T_1}$ .
So the answer is (d) $m\ln\left(\frac{T_2}{T_1}\right)$ .

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