1. Example 1: How much heat (in calories) is needed to raise 20
grams of water from 5 o
C to 40 o
C?
Q = m c (Tf—Ti)
Q = x
m = 20 gr
c = 1 cal/go
C (this is the specific heat of water)
Tf = 40 o
C
Ti = 5 x = (20 g) (1 cal/go
C)(400
C) -5 0
C) = 700 cal
Example 2:
How much heat (in calories) is needed to raise 140 grams of water from
20 o
C to 25 o
C?
Q = mc(Tf—Ti)
Q = x
m = 140 grams
c = 1 cal/go
C (this is the specific heat of water)
Tf = 25 o
C
Ti = 20 o
C
x = (140 g) (1 cal/g o
C)(25-20 0
C) = 700 cal

2. Example 3:
How much heat (in calories) is needed to raise 250 grams of water from
80 o
C to 87 o
C?
Q = mc(Tf—Ti)
Q = x
m = 250 grams
c = 1 cal/go
C (this is the specific heat of water)
Tf = 87 o
CTi = 80 o
C
x = (250 g) (1 cal/goC)(87—80 0
C) = 1750 cal
Example 4:
How many calories are needed to raise50 grams of iron from55 oC to
200 oC? The specific heat capacity of iron is 0.11 cal/go
C.
Q = mc(Tf—Ti)
Q = x
m = 50 grams
c = 0.11 cal/goC
Tf = 200 o
C
Ti = 55 o
C
x = (50 g) (0.11 cal/go
C)(200–55 0
C) = 797.5 cal

3. Example 5:
How many grams of aluminum can be heated from90 o
C to 120 o
Cif 500
calories are applied? The specific heat of aluminum is 0.21 cal/go
C.
Q = mc(Tf—Ti)
Q = 500 calories
m = x grams
c = 0.21 cal/go
C
Tf = 120 o
C
Ti = 90 o
C
500 cal= (x) (0.21 cal/go
C)(120—90 0
C)
x = 79.4 grams
Example 6:
What is the specific heat capacity of a substanceif 400 calories cause 25
grams of it to go from 60 o
Cto 190 o
C?
Q = mc(Tf—Ti)
Q = 400 calories
m = 25 grams
c = x cal/go
C
Tf = 190 o
C
Ti = 60 o
C
400 calories = (25 g) (x)(130 0
C)
x = 0.123 cal/go
C

4. Example 7:
What is the final temperature if 500 calories are applied to 40 grams of
copper at 20 o
C? The specific heat capacity of copper is 0.092 cal/go
C.
Q = mc(Tf—Ti)
Q = 500
m = 40 grams
c = 0.092 cal/go
C
Tf = x
Ti = 20 o
C
500 cal = (40 g) (0.092 cal/goC)(x–20 0C)
x = 156 o
C
Example 8:
What was the initial temperature if 250 calories were applied to 100
grams of gold and the final temperature of the gold was 175 o
C?
The specific heat capacity of gold is 0.031 cal/go
C.
Q = mc(Tf—Ti)
Q = 250 calories
m = 100 grams
c = 0.031 cal/go
C
Tf = 175 o
C
Ti = x o
C
250 cal= (100 g) (0.031 cal/go
C)(175—x0C)
x = 94 o
C