a calorimeter contains a mixture of 250gm of water and 200gm of ice at 0C.the water equivalent of calorimeter is 60g..Now 300gm of steam at 100C is passed through mixture,calculate final temperature of mixture.

latent heat of steam=536cal/gm and latent heat of ice=80cal/gm

Hi, For any calorimetry experiment, heat lost = heat gained. Here, ms(for calorimeter)= 60cal/K(orC) , Now, heat gained = mass of ice x latent heat of fusion + mass of water x specific heat capacity of water x (T-0) + heat gained by calorimeter and heat lost = mass of steam x latent heat of vaporisation + mass of water formed x sp. heat capacity of water x (100-T) Equating both: 200 x 80 + 450 x 4.2 x (T-0)+ 60(T-0) = 300 x 536 + 300 x 4.2 (100-T) Note that: i)we get 450 g of water when ice melts and 300 g of water when steam cools down ii)calorimeter is also at 0 C initially, iii)sp. heat capacity of water= 4.2 cal/g C Solving the above equation: T = 84.3 C (approx.) ANS. (If the whole of steam does not convert into water or whole of ice does not melt, you may get ambiguous answers, like T=-ve or T> 100, which are not possible. You are correct only 0