An amount of ice of mass 10⁻³ kg and temperature -10 °C is transformed to vapour at a temperature of 110 °C by applying heat. The total amount of heat 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…

Question

An amount of ice of mass 10⁻³ kg and temperature -10 °C is transformed to vapour at a temperature of 110 °C by applying heat. The total amount of heat 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)

Accepted Answer

(1) 3043 J. Given: m=10⁻³ kg. Five legs in order, sensible heat before each latent heat. Warm ice -10→0 °C: 10⁻³(2100)(10)=21 J. Melt it at 0 °C: 10⁻³(3.35×10⁵)=335 J. Warm water 0→100 °C: 10⁻³(4180)(100)=418 J. Boil it at 100 °C: 10⁻³(2.25×10⁶)=2250 J. Superheat the steam 100→110 °C: 10⁻³(1920)(10)=19.2 J. Q=21+335+418+2250+19.2=3043.2 J≈3043 J.

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