Density of water at 4^ C and 20^ C are 1000 kg/m^3 and 998 kg/m^3… — Solution | JEE Physics

MCQEasyJEE 2026First Law & Internal Energy

Density of water at $4^\circ\text{C}$ and $20^\circ\text{C}$ are $1000\,kg/m^3$ and $998\,kg/m^3$ 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 \hspace{1cm J.

(Specific heat capacity of water $= 4.2\,J\,g^{-1}K^{-1}$ and atmospheric pressure $=10^5\,Pa$ )

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

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Mass of water $m = 4\,\text{kg}$ , specific heat capacity $c = 4200\,\text{J kg}^{-1}\text{K}^{-1}$ , temperature change from $4^\circ\text{C}$ to $20^\circ\text{C}$ , pressure $P = 10^5\,\text{Pa}$ , density at $4^\circ\text{C}$ is $1000\,\text{kg/m}^3$ and density at $20^\circ\text{C}$ is $998\,\text{kg/m}^3$ .

Find: Increase in internal energy $\Delta U$ .

At constant pressure,

\Delta U = Q - W

where $Q = mc\Delta T$ and $W = P\Delta V$ .

First, calculate heat supplied:

Q = mc\Delta T

Q = 4 \times 4200 \times (20-4)

Q = 268800\,\text{J}

Now calculate the initial and final volumes:

V_1 = \frac{4}{1000} = 0.004\,\text{m}^3

V_2 = \frac{4}{998} = 0.004008\,\text{m}^3

Hence,

\Delta V = V_2 - V_1 = 8 \times 10^{-6}\,\text{m}^3

Work done at constant pressure is:

W = P\Delta V = 10^5 \times 8 \times 10^{-6} = 0.8\,\text{J}

Therefore,

\Delta U = Q - W

\Delta U = 268800 - 0.8 = 268799.2\,\text{J}

The increase in internal energy is $268799.2\,\text{J}$ . Therefore, the correct option is D.

Common mistakes

Using $\Delta U = Q$ directly without subtracting expansion work. This is wrong because the water expands at atmospheric pressure, so a small amount of heat goes into doing work. Use $\Delta U = Q - P\Delta V$ instead.
Using inconsistent units for specific heat capacity. The given value is $4.2\,\text{J g}^{-1}\text{K}^{-1}$ , which must be converted consistently when mass is taken in kilograms. Either use $4\,\text{kg}$ with $4200\,\text{J kg}^{-1}\text{K}^{-1}$ or use grams throughout.
Calculating volume change incorrectly from density data. Since $V = \frac{m}{\rho}$ , the initial and final volumes must be found separately before subtracting. Do not subtract densities first and treat that as volume change.

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Answer

Step-by-step solution

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