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Heat and Thermodynamics question

2011 · Shift 0 · Q68

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Heat and Thermodynamics
2011 · Shift 0 · Q68

Heat and Thermodynamics question

2011 · Shift 0 · Q68

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1

100g

of water is heated from

{30^ \circ }C

to

{50^ \circ }C

. Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is

4184J/kg/K

):

A
$8.4kJ$
B
$84kJ$
C
$2.1kJ$
D
$4.2kJ$

View written solutionFree

Correct answer: A

Given data

Mass of water: $100\text{ g} = 0.1\text{ kg}$
Initial temperature: $30^\circ \text{C}$
Final temperature: $50^\circ \text{C}$
Specific heat of water: $c = 4184\,\text{J/kg/K}$

Temperature rise

\Delta T = 50 - 30 = 20\,\text{K}

Relation for change in internal energy

Since the slight expansion of water is ignored, no significant work is done. Hence,

\Delta U \approx Q = mc\Delta T

Substitute the values

\Delta U = (0.1)(4184)(20)

First calculate:

4184 \times 20 = 83680

Then,

0.1 \times 83680 = 8368\,\text{J}

Convert to kJ

8368\,\text{J} = 8.368\,\text{kJ} \approx 8.4\,\text{kJ}

Evaluate options

A: $8.4\,\text{kJ}$ ✅
B: $84\,\text{kJ}$ ❌
C: $2.1\,\text{kJ}$ ❌
D: $4.2\,\text{kJ}$ ❌

Therefore, the correct answer is A.

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