1A: How many calories will it take to raise the temperature of a 39 g gold chain from 20Â°C to 110Â°C? _________ cal 1B: How many calories would it take to raise the temperature of a 415 g aluminum pan from 293 K to 384 K? _________ cal 1C: 75 grams of water at 90Â°C are mixed with an equal amount of water at 10Â°C in a completely insulated container. The final temperature of the water is 50Â°C. (a) How much heat is lost by the hot water? __________ cal (magnitude only) (b) How much heat is gained by the cold water? __________cal (magnitude only) (c) What happens to the total amount of internal energy of the system? A: remains unchanged B: Halves C: Doubles 1D: A kettle containing 2 kg of water has just reached its boiling point. How much energy, in joules, is required to boil the kettle dry? __________ J Solution The specific heat of gold is 0.031 calories/gramÂ°C The calories it will take to raise the temperature of a 39 g gold chain from 20Â°C to 110Â°C = 0.031*39*(110-20) = 108.81 Cal The specific heat of Al is 0.22 calories/gramÂ°C 293 K = 20C 384 K = 111 C The calories it will take to raise the temperature of a 415 g aluminum pan from 20Â°C to 111Â°C = 0.22*415*(111-20) = 8308.3 Cal The specific heat capacity of water is 4200 J kg-1 Â°C-1 Heat lost by hot water = 0.075*4200*(90-50) = 12600 J = 3011.4 Cal Heat gained by cold water = Heat lost by hot water = 3011.4 Cal The total amount of interanl energy remains unchanged since it is insulated. The heat of vaporization is 2257 J/g. Therefore your answer is 2,257,000*2 J = 4514000 J .