Specific Heat Worksheet #2
(Answer Key)
1.
Brass is an alloy made from copper and zinc. A 0.66 kg sample of brass at 98.6oC
is dropped into 2.33 kg of water at 4.6oC. If the equilibrium temperature is 7.0oC,
what is the specific heat capacity of brass.
.66kg x(1000g/1kg) x (c) x [98.6 -7] = 2.33kg x (1000g/1kg) x 4.18 J/goC x [7 - 4.6]
c = .387J/groC
A 0.012kg cube of ice at 0.0oC is added to 0.459 kg of soup at 80.4oC. Assuming
that the soup has the same specific heat capacity as water, find the final
temperature of the soup after the ice has melted. The molar heat of fusion is
6.01 kJ/mol.
0.12kg x (1000g/1kg) x(1mol/18g)x 6.01kJ/mol = energy absorbed by the ice cube =
40.06kJ
2.
40.06kJ came from the hot soup thus 40.06kJ = .459kg x (4.18J/groC) x (tf - 80.4oC)
convert all to Joules thus 40.06kJ x 1000J/1kJ = 40,060 J which means the soup lost this
much energy thus it is a negative value.
-40,060 J = .459kg x (1000gr/1 kg) x (4.18 J/groC) x (tf - 80.4oC)
-20.87oC = Tf - 80.4oC
Tf = 59.53oC
3.
A 25.4 g ring that appears to be silver is heated to a temperature of 85oC, and then
is placed in a calorimeter containing .05 kg of water @ 24.8oC. The calorimeter
is not perfectly insulated, however, so that 0.11 kJ of energy is transferred by heat
to the surroundings by the time a temperature of 25oC reached. From the info
provided, determine the specific heat capacity of the ring.
.05kg x (1000g/ 1kg) x (4.18 J/groC) x [ 24.8oC - 25oC] = 25.4 g x ( c ) x [25 - 85] + .11kJ x 1000 J/1kJ
-418 J = = 25.4 g x ( c ) x [25 - 85] + 110 J
c = .346 J/groC
4.
A sheet of gold weighing 10.0 grams and at a temperature of 18.0oC is place flat
on a sheet of iron weighing 20.0 grams and at a temperature of 55.6oC. What is
the final temperature of the combined metals? Assume that no heat is lost to the
surroundings. (Hint: The heat gained by gold must be equal to the heat lost by the
iron) ( ciron = ..444 J/goC and cgold = .129 J/goC)
10 gr x (.129 J/groC) x [Tf - 18.0oC] = -20.0 gr x (.444 J/groC) x [Tf - 55.6]
negative sign is due to loss of energy
1.29 J/oC [Tf - 18oC] = -8.88 J/oC [Tf - 55.6oC]
.145[Tf - 18] = -Tf + 55.6
.145Tf - 2.61 = -Tf + 55.6
1.145Tf = 58.21oC
Tf = 50.84oC