The document provides 8 examples of using the heat transfer equation Q=mcΔT to calculate heat transfer values for various substances. Specifically, it shows how to calculate the heat (in calories) needed to change the temperature of given masses of substances like water, iron, aluminum, and copper between provided initial and final temperatures. It also shows how to calculate initial or final temperatures when other values like mass, heat input, and specific heat are known.

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