JEE Main 2025 Jan 22 Shift 1, Physics Q6: Change of State And Calorimetry

JEE Main2025Jan 22, Shift 1Physics

An amount of ice of mass $1 0^{- 3}$ kg and temperature $- 1 0^{\circ}$ C is transformed to vapour of temperature $11 0^{\circ}$ by applying heat. The total amount of work required for this conversion is,

(Take, specific heat of ice $= 2100$ Jkg $^{- 1}$ K $^{- 1}$ , specific heat of water $= 4180$ Jkg $^{- 1}$ K $^{- 1}$ , specific heat of steam $= 1920$ Jkg $^{- 1}$ K $^{- 1}$ , Latent heat of ice $= 3.35 \times 1 0^{5}$ Jkg $^{- 1}$ and Latent heat of steam $= 2.25 \times 1 0^{6}$ Jkg $^{- 1}$ )

A
3022 J
B
3043 J
C
3003 J
D
3024 J

Solution

Sum the five stages with $m = 1 0^{- 3}$ kg:

(1) Warm ice from $- 1 0^{\circ}$ to $0^{\circ}$ C: $Q_{1} = m s_{i ce} Δ T = 1 0^{- 3} \times 2100 \times 10 = 21$ J.

(2) Melt ice at $0^{\circ}$ C: $Q_{2} = m L_{f} = 1 0^{- 3} \times 3.35 \times 1 0^{5} = 335$ J.

(3) Warm water from $0^{\circ}$ to $10 0^{\circ}$ C: $Q_{3} = 1 0^{- 3} \times 4180 \times 100 = 418$ J.

(4) Vaporise water at $10 0^{\circ}$ C: $Q_{4} = 1 0^{- 3} \times 2.25 \times 1 0^{6} = 2250$ J.

(5) Heat steam from $10 0^{\circ}$ to $11 0^{\circ}$ C: $Q_{5} = 1 0^{- 3} \times 1920 \times 10 = 19.2$ J.

Total $= 21 + 335 + 418 + 2250 + 19.2 = 3043.2$ J $\approx 3043$ J.

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JEE Main 2025 Jan 22 Shift 1, Physics Q6: Change of State And Calorimetry

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