1A: How many calories will it take to raise the temperature of a 39 g gold chain from
20°C to 110°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 90°C are mixed with an equal amount of water at 10°C in a
completely insulated container. The final temperature of the water is 50°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/gram°C
The calories it will take to raise the temperature of a 39 g gold chain from 20°C to 110°C =
0.031*39*(110-20) = 108.81 Cal

The specific heat of Al is 0.22 calories/gram°C
293 K = 20C
384 K = 111 C
The calories it will take to raise the temperature of a 415 g aluminum pan from 20°C to
111°C = 0.22*415*(111-20) = 8308.3 Cal
The specific heat capacity of water is 4200 J kg-1 °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