JEE Main 2025 Jan 23 Shift 2, Physics Q15: Laws of Thermodynamics

JEE Main2025Jan 23, Shift 2Physics

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 $1 J k g^{- 1} K^{- 1}$ , is :

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

Solution

For a reversible (slow) heating, $d Q = m s d T$ and entropy change is

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

With $s = 1$ in the given units, $Δ S = m ln (\frac{T _{2}}{T _{1}}) .$

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JEE Main 2025 Jan 23 Shift 2, Physics Q15: Laws of Thermodynamics

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