More practice questions

4. Q4. How much heat turns 0.20 kg of ice at −10 °C into water at 20 °C? Use NCERT’s values: ice 2060 J kg^⁻¹ K^⁻¹, water 4186 J kg^⁻¹ K^⁻¹, L_f = 3.33 × 10^⁵ J kg^⁻¹.

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   Warm the ice: 0.20 × 2060 × 10 = 4120 J. Melt it: 0.20 × 3.33 × 10^⁵ = 66 600 J. Warm the water: 0.20 × 4186 × 20 = 16 744 J. Total 87 464 J, about 8.7 × 10^⁴ J.

5. Q5. 10 g of steam at 100 °C is passed into 0.20 kg of water at 20 °C. Ignoring the container and heat loss, find the final temperature.

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   Heat given by the steam: 0.010 × 22.6 × 10^⁵ + 0.010 × 4186 × (100 − T). Heat taken by the water: 0.20 × 4186 × (T − 20). Setting them equal: 22 600 + 4186 − 41.86 T = 837.2 T − 16 744, so 879.06 T = 43 530 and T ≈ 49.5 °C.

6. Q6. Why can solid carbon dioxide (dry ice) not melt into a liquid in an open room?

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   Its triple point is at 5.11 atm (NCERT figure, page 211). Below that pressure, liquid carbon dioxide cannot exist, so at 1 atm the solid turns straight into gas (it sublimes) at −78.5 °C.

7. Q7. A pure substance of fixed mass absorbs heat at constant power, with negligible losses, while changing phase at fixed pressure. On its temperature–time graph, melting lasts 2.0 min and boiling lasts 13.6 min. Find L_v/L_f.

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   With constant power P, the heat taken in is Q = Pt, and during each change of state Q = mL for the same mass. So L_v/L_f = 13.6/2.0 = 6.8, the value for water.