Specific Heat Capacity
Target 3: Calculate the specific heat capacity of a substance
Specific heat is a physical property of matter. The temperature of a substance is a measure of the motion (kinetic
energy) of the molecules. It is a directly proportional relationship; the greater the motion (kinetic energy) the higher the
temperature. Motion requires energy. Thus, the more energy matter has the higher temperature it has. Typically this
energy is supplied by heat. Heat loss or gain by matter is equivalent energy loss or gain. The higher the specific heat
capacity of a substance, the more energy it takes to change its temperature and the slower it changes temperature. The
lower the specific heat capacity, the less heat energy it takes to change its temperature and the faster it will do so.
Us the formula Q = cmΔT to solve for the variable indicated.
Q = Heat (Joules)
c = Specific heat
m = Mass (grams) kg
T = Temperature (°C)
Corrections in RED
ΔT = Change in temperature (°C)
BASIC PROFICIENT
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 of this object?
Formula (1 pt.)
C = Q/mΔT
Variables (1 pt.)
C=?
Final Answer (5 pts.)
m = 125g
2006.40 J/kg°C
Q = 300 cal
ΔT = 5°C
300 cal
4.18 J
1
1000 g
1
1
1 cal
125 g
1 kg
5°C
Work
(3 pts.)has a specific heat of 0.215 cal/g oC. How much heat is lost when a piece of aluminum with a mass of
2. Aluminum
24g cools from a temperature of 415.0oC to a temperature of 22.0oC?
Formula (1 pt.)
Q = cmΔT
Variables (1 pt.)
Final Answer (5 pts.)
C = 0.215 cal/g°C
m = 24g
ΔT = 5°C
Q=?
107.84 J
.215 cal
1000 g
4.18 J
24 g
1 kg
5°C
g°C
1 kg
1 cal
1
1000 g
1
Work (3 pts.)
1
PROFICIENT
3. 850J of heat are applied to a 250g sample of liquid water with an initial temperature of 13.0oC. Find the change
in temperature.
Formula (1 pt.)
ΔT = Q/cm
Variables (1 pt.)
Final Answer (5 pts.)
C = 1 cal/g°C
m = 250g
ΔT = ?
Q = 850J
0.81 °C
Work (3 pts.)
850 J
g°C
1 cal
1 kg
1
1000 g
1
1 cal
4.18 J
1000 g
250 g
1 kg
4. How many calories does it take to heat a 25g piece of iron (c= .49 cal/g oC) is heated from room temperature
(24oC) to 165oC?
Formula (1 pt.)
Q = cmΔT
Variables (1 pt.)
C = 0.49 cal/g°C
m = 25g
ΔT = 141°C
Final Answer (5 pts.)
7219.91 J
Q=?
0.49 cal
1000 g
4.18 J
25 g
1 kg
141°C
g°C
1 kg
1 cal
1
1000 g
1
Work (3 pts.)
A 120 gram piece of aluminum (cp = .215 cal/g°C) is at 90°C. It is placed in a beaker that contains 35 grams of 25°C
water. At what temperature will they come to thermal equilibrium?
Qw = Qal
cmΔt = cmΔt
(1)(35)(Tf – 25) = (.215)(120)( Tf – 90)
35(Tf – 25) = 25.8( Tf – 90)
35Tf – 875 = 25.8Tf – 2322
35Tf – 875 = -25.8Tf + 2322
60.8Tf – 875 = 2322
{swap signs}
Combine like terms
60.8Tf = 3197
Tf = 52.58°C
2
3