5 g of water at 30 ° C and 5 g of ice at - 20 ° C are mixed together in a calorimeter. Find the final temperature of the mixture. Assume water equivalent of calorimeter to be negligible, specific heats of ice and water are 0 . 5 and 1 cal g - 1 ° C - 1 , and latent heat of ice is 80 cal g - 1 .

Step-by-step solution

Here ice will absorb heat while hot water will release it. So if T is the final temperature of the mixture, heat given by water Q 1 = m c Δ T = 5 × 1 × 3 0 - T And heat absorbed by the ice Q 2 = 5 × 1 2 0 - - 2 0 + 5 × 8 0 + 5 × 1 T - 0 So, by the principle of calorimetry Q 1 =   Q 2 , i.e., 150   -   5 T   =   450   +   5 T   T = -   30   ° C Which is impossible as a body cannot be cooled to a temperature below the temperature of