An amount of ice of mass kg and temperature –10°C is transformed to vapour of temperature 110°C by applying heat. The total amount of work required for this conversion is, (Take, specific heat of ice = , specific heat of water = , specific heat of steam

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Solution

Q.
An amount of ice of mass $10^{- 3}$ kg and temperature –10°C is transformed to vapour of temperature 110°C by applying heat. The total amount of work required for this conversion is, (Take, specific heat of ice = $2100 J {kg}^{- 1} K^{- 1}$ , specific heat of water = $4180 J {kg}^{- 1} K^{- 1}$ , specific heat of steam = $1920 J {kg}^{- 1} K^{- 1}$ , Latent heat of ice = $3.35 \times 10^{5} J {kg}^{- 1}$ and Latent heat of steam = $2.25 \times 10^{6} J {kg}^{- 1}$ ) [2025]

1
3022 J

2
3043 J

3
3003 J

4
3024 J

Ans.
(2)

First, Ice at –10°C $\overset{△ Q_{1}}{\to}$ Ice at 0°C $\overset{△ Q_{2}}{\to}$ Water at 0°C. Then, Water at 0°C $\overset{△ Q_{3}}{\to}$ Water at 100°C $\overset{△ Q_{4}}{\to}$ Vapour at 100°C $\overset{△ Q_{5}}{\to}$ Vapour at 110°C

$△ Q_{1} = m \times S_{I} \times △ T = 10^{- 3} \times 2100 \times 10 = 21 J$

$△ Q_{2} = m \times L_{f} = 10^{- 3} \times 3.35 \times 10^{5} = 335 J$

$△ Q_{3} = m \times S_{w} \times △ T = 10^{- 3} \times 4180 \times 100 = 418 J$

$△ Q_{4} = m \times L_{v} = 10^{- 3} \times 2.25 \times 10^{6} = 2250 J$

$△ Q_{5} = m \times S_{v} \times △ T = 10^{- 3} \times 1920 \times 10 = 19.2 J$

$△ Q_{net} = 3043.2 J$

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