Name: _________________________________________
Period: ______
Target: Calculate the heat transfer to/from an substance using the
formula Q = cm∆t
1. When 300cal of energy is lost from a 125g object, the temperature decreases from 45°C to
40°C. What is the specific heat capacity of the object?
Formula (1 pt.)
c = Q/mΔT
Variables (3 pt.)
Q = 300 cal
m = 125g
Final Answer (6 pts.)
0.48 cal/g°C
ΔT = 5 °C
Work (3 pts.)
300 cal
1
1
1
125g
5°C
2. Aluminum has a specific heat of 0.22 How much heat is lost when a piece of aluminum with a
mass of 14g cools from a temperature of 315.0oC to a temperature of 22.0oC?
Formula (1 pt.)
Variables (3 pt.)
m = 14g
Q=?
Q = cmΔT
c = 0.22 cal/g oC
ΔT = 293 oC
Final Answer (6 pts.)
902.44 cal
Work (3 pts.)
.022 cal
14 g
293 oC
g oC
1
1
3. 8500J of heat is applied to a 150g sample of liquid water with an initial temperature of 13.0oC.
Find the change in temperature.
Formula (1 pt.)
ΔT = Q/cm
Variables (3 pt.)
Q = 8500 J
m = 150g
Final Answer (6 pts.)
13.56 oC
o
c = 1.00 cal/g C
Work (3 pts.)
8500 j
1 cal
g oC
1
1
4.18 j
1 cal
150 g
1
Name: _________________________________________
Period: ______
4. How many calories does it take to heat a 25g piece of iron (c = 0.49 ca/g°C) from room
temperature (24°C) to 165°C?
Formula (1 pt.)
Variables (3 pt.)
m = 25 g
Q=?
Q = cmΔT
c = 0.49 cal/g°C
ΔT = 141 °C
Final Answer (6 pts.)
1727.25 cal
Work (3 pts.)
0.49 cal
25 g
141 °C
g°C
1
1
5. A 1000g sample of metal with a specific heat capacity of 0.112 cal/g°C is heated to 100°C and
then cooled to 65°C. How many calories were “lost”?
Formula (1 pt.)
Variables (3 pt.)
m = 1000 g
Q=?
Q = cmΔT
c = 0.112 cal/g°C
ΔT = 35 °C
Final Answer (6 pts.)
3920.00 cal
Work (3 pts.)
0.112 cal
1000 g
35 °C
g°C
1
1
6. 8000 calories are applied to a 250g sample of liquid water and the temperature increase by
32°C. What is the specific heat capacity of water?
Formula (1 pt.)
c = Q/mΔT
Variables (3 pt.)
Q = 8000 cal
m = 250 g
ΔT = 32 °C
c=?
Final Answer (6 pts.)
1.00 cal/g°C
Work (3 pts.)
8000 cal
1
1
1
250 g
32 °C
2