Concept: According to the principle of the calorimeter, Heat gain by ice = Heat lost by water + Heat lost by the calorimeter mihl + mwcwdt = mcwdt + ccdt Where, hl = Latent heat; mi, mw = Mass of ice and water; cw,cc = Specific heat of water and calorimeter; Calculation: Given: mi = 10 gm; mw = 10 gm; cc = 5 cal/°C; hl = 80 cals/gm; We know that cw = 1 cal/gm °C Using, mihl + mwcwdt = mcwdt + ccdt 10 × 80 + 10 × 1 × (T - 0) = 10 × 1 × (25 - T) + 5(25 - T) 800 + 10T = 250 - 10T + 125 - 5T 800 + 10 T = 375 - 15 T T = - 19°C Which is impossible So, let's check how much heat can water and calorimeter loose Max. heat which can be lost by water = 10 × 1 × (25 - 0) = 250 cal Max. heat which can be lost by calorimeter = 5 × (25 - 0) = 125 cal 10 × 1 × (25 - 0) = 250 cal Total heat which can be lost = 250 + 125 = 375 cal To convert ice into water, we require = 10 × 80 = 800 cal Hence, ice will not melt and the final temperature of system will be 0° C