Thermal Properties of Matter

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 =3.35 105 J kg-1 and Latent heat of steam =2.25 106 J kg-1 )

1. 3022 J
2. 3043 J
3. 3003 J
4. 3024 J

Answer

(2) 3043 J

Solution

Five stages: warm ice, melt, warm water, boil, warm steam (m=10-3 kg).

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Q1=10-3 2100 10=21 J

Q2=10-3 3.35 105=335 J

Q3=10-3 4180 100=418 J

Q4=10-3 2.25 106=2250 J

Q5=10-3 1920 10=19.2 J

Q=21+335+418+2250+19.2 3043 J

Answer: (2)