Heat absorbed by the ice when it changes into water (Q 1) = 1×(3.36 × 10 5) J Therefore, ice will first change into water as less heat is required for this and there will be equilibrium between steam and water. We can observe that the latent heat of fusion of ice (3.36 × 10 5 J kg −1) is smaller that latent heat of vaporisation of water (2.26 × 10 6 ). Latent heat of vaporisation of water = 2.26 × 10 6 J kg −1

Latent heat of fusion of ice = 3.36 × 10 3 J kg −1 Heat released when temperature of 200 ml changes from 10 oC to 0 oC is given by Thus, the thermal equilibrium will be attained at 0 o C.Įquilibrium temperature of the cube and the drink = 0☌ Since the heat required for melting the four cubes of the ice is greater than the heat released by water ( H i > H w ), some ice will remain solid and there will be equilibrium between Let us first check the heat released when temperature of 200 ml changes from 10 oC to 0 oC. Latent heat of fusion of ice, L i = 3.4 × 10 5 J kg −1 Volume of each ice cube = (2 × 2 × 2) = 8 cm 3 Using the values of MC A and MC C, we get Heat gained by liquid A = Heat lost by liquid C Let the temperature of the mixture be T. Then, Heat gained by liquid B = Heat lost by liquid C Heat gained by Liquid A = Heat lost by liquid B Let the mass of the mixtures be M and the specific heat capacities of the liquids A, B and C be C A, C B, and C C, respectively.Īccording to the principle of calorimetry, when A and B are mixed, we get

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