JEE Main 2026 Jan 24 Shift 1, Physics Q4: Laws of Thermodynamics

JEE Main2026Jan 24, Shift 1Physics

Density of water at $4^{\circ} C$ and $2 0^{\circ} C$ are 1000 kg/m $^{3}$ and 998 kg/m $^{3}$ respectively. The increase in internal energy of 4 kg water when it is heated from $4^{\circ} C$ to $2 0^{\circ} C$ is____J.

(Specific heat capacity of water = 4.2 J/kg. and 1 atmospheric pressure = $1 0^{5}$ Pa)

A
$315826.2$
B
$234699.2$
C
$258700.8$
D
$268799.2$

Solution

(Note: "Specific heat capacity 4.2 J/kg" in the problem should read 4.2 J/g $\cdot$ K = $4200$ J/kg $\cdot$ K.)

Heat added:

$Q = m c Δ T = 4 \times 4200 \times 16 = 268800$ J.

Work done by water against atmosphere due to thermal expansion:

$Δ V = m (\frac{1}{ρ _{f}} - \frac{1}{ρ _{i}}) = 4 (\frac{1}{998} - \frac{1}{1000}) = 4 \times \frac{2}{998 \times 1 0 ^{3}} = \frac{8}{9.98 \times 1 0 ^{5}}$ m $^{3}$ .

$W = P Δ V = 1 0^{5} \times \frac{8}{9.98 \times 1 0 ^{5}} \approx 0.8$ J.

First law:

$Δ U = Q - W = 268800 - 0.8 = 268799.2$ J.

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JEE Main 2026 Jan 24 Shift 1, Physics Q4: Laws of Thermodynamics

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