Consider 50 g of hot water at 80°C poured into a cavity in a very large block of ice at 0°C. What will be the final temperature of the water in the cavity? Show that 50 g of ice must melt in order to cool the hot water down to this temperature.
Solution 4P Part 1 Since the ice block is very large, all the heat released by water will be used to melt some amount of the ice surrounding the water. When the temperature of the water comes down to 0°C, both the water and the ice will have same temperature and hence there will be no exchange of heat. So the final temperature of the block and water will be 0°C. Part 2 Step 1 Let us consider the mass of the ice that will melt is . ice The latent heat of melting of ice is given by L= 80 cal/g ice So the heat required to melt iceamount of ice is given by Q ice= m ice ice mice0 cal/g) Step 2 Heat released by the hot water is given by Q water= m water water water Where m = 50 g is the mass of the water, c = 1 cal/g°C is the specific heat of the water and water water T water (80°C 0°C) = 80°C is the change in the temperature of the water. Hence we have Q = (50 g)(1 cal/g°C)(80°C) = 4000 cal water