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 $1 \mathrm{Jkg}^{-1} \mathrm{~K}^{-1}$, is :

(D) $m \ln\left(\frac{T_2}{T_1}\right)$

(C) $m \ln \left(\frac{T_1}{T_2}\right)$

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Heat & Thermodynamics

Water of mass

m

gram is slowly heated to increase the temperature from

T_{1}

T_{2}

. The change in entropy of the water, given specific heat of water is

1 Jkg^{- 1} K^{- 1}

, is :

(A)zero
(B) $m (T_{2} - T_{1})$
(C) $m ln (\frac{T _{1}}{T _{2}})$
(D) $m ln (\frac{T _{2}}{T _{1}})$ ✓ Correct

Answer: (D)

Solution

For a reversible process,

d S = \frac{d Q}{T}

. Heating the water gives

d Q = m s d T

d S = \frac{m s d T}{T}

Integrating from

T_{1}

T_{2}

Δ S = m s \int_{T_{1}}^{T_{2}} \frac{d T}{T} = m s ln (\frac{T _{2}}{T _{1}})

With specific heat

s = 1 J kg^{- 1} K^{- 1}

, this reduces, in terms of the given mass

m

, to

Δ S = m ln (\frac{T _{2}}{T _{1}})

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option

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