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Thermodynamics Part 5

Section: Physics

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Question : 26 of 86

Marks: +1, -0

Density of water at

4^{\circ}\text{C}

20^{\circ}\text{C}

1000\;\text{kg/m}^3

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}

20^{\circ}\text{C}

\_\_\_\_

(specific heat capacity of water

=4.2\;\text{J/kg}

. and 1 atmospheric pressure

=10^5\;\text{Pa}

[24 jan 2026 Shift 1]

268799.2
315826.2
234699.2
258700.8

Solution:

Concept:

Use first law of thermodynamics:

\Delta U = Q - W

.

Q

is heat supplied at constant pressure and

W

is work done due to expansion.

Explanation:

Heat supplied:

Q = m c \Delta T

m = 4

kg,

c = 4.2 \times 10^3

J kg

^{-1}

K

^{-1}

,

\Delta T = 20-4 = 16

K.

Q = 4 \times 4.2 \times 10^3 \times 16 = 268800

J.

Initial volume:

V_1 = \frac{m}{\rho_1} = \frac{4}{1000} = 0.004

m

^3

.

Final volume:

V_2 = \frac{m}{\rho_2} = \frac{4}{998} \approx 0.004008016

m

^3

.

Change in volume:

\Delta V = V_2 - V_1 \approx 8.016 \times 10^{-6}

m

^3

.

Work done:

W = P \Delta V = 10^5 \times 8.016 \times 10^{-6} = 0.8016

J.

Apply first law:

\Delta U = Q - W = 268800 - 0.8016 = 268799.1984

J.

Round to one decimal:

268799.2

J.

Answer:

Option A: 268799.2 J

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