Density of water at 4°C and 20°C are 1000 kg/m³ and 998 kg/m³ respectively. The increase in internal energy of 4 kg of water when it is heated from 4°C to 20°C is ______ J. (specific heat capacity of water = 4.2 kJ/kg K and 1 atmospheric pressure = 10⁵ Pa) (a) 268799.2 (b) 315826.2 (c) 234699.2 (d) 258700.8

Answer: (a). First law: Δ U = Q - PΔ V. Q = mcΔ T = 4 × 4200 × 16 = 268800 J Δ V = 4/998 - 4/1000 ≈ 8.016 × 10⁻⁶ m³, so PΔ V = 10⁵ × 8.016 × 10⁻⁶ ≈ 0.8 J Δ U = 268800 - 0.8 = 268799.2 J So the answer is (a) 268799.2.

Density of water at 4°C and 20°C are 1000 kg/m³ and 998 kg/m³… | JEE Main 2026 (24 Jan, Shift 1)

Q10JEE Main 2026 · 24 JAN · SHIFT 1Difficult

Calorimetry, Latent Heat & Mechanical Equivalent of Heat

Density of water at $4^{\circ}\text{C}$ and $20^{\circ}\text{C}$ are $1000\text{ kg/m}^{3}$ and $998\text{ kg/m}^{3}$ respectively. The increase in internal energy of 4 kg of water when it is heated from $4^{\circ}\text{C}$ to $20^{\circ}\text{C}$ is ______ J.

(specific heat capacity of water $= 4.2\text{ kJ/kg K}$ and 1 atmospheric pressure $= 10^{5}\text{ Pa}$ )

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Answer: (a)

First law: $\Delta U = Q - P\Delta V$ .
$Q = mc\Delta T = 4 \times 4200 \times 16 = 268800\ \text{J}$
$\Delta V = \dfrac{4}{998} - \dfrac{4}{1000} \approx 8.016 \times 10^{-6}\ \text{m}^3$ , so $P\Delta V = 10^5 \times 8.016 \times 10^{-6} \approx 0.8\ \text{J}$
$\Delta U = 268800 - 0.8 = 268799.2\ \text{J}$
So the answer is (a) 268799.2.

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