Name
Date
Pd
Unit 3 Worksheet 4 – Quantitative Energy Problems
Part 2
Energy constants (H2O)
Q  m  c  T(C)
Heat of fusion (melting or freezing) Hf
Q  m  H v or m  H f
Heat of vaporization (evaporating or condensing) Hv
Heat capacity (c) of solid water
Heat capacity (c) of liquid water

For each of the problems sketch a warming or cooling curve to help you decide which equation(s) to
use to solve the problem. Keep a reasonable number of sig figs in your answers.
334 J/g
2260 J/g
2.1 J/g˚C
4.18 J/g˚C
T (°C)
1. How much energy must be absorbed by a 150 g sample of ice at 0.0 ˚C that melts and then warms
to 25.0˚C?
2) liquid H2O
warms up
25
1) ice melts
-25
Q1  m  H f
Q2  m  c  T
Q1  150 g  334 J g
Q1  50,100 J
Q2  150 g  4.18 J gC  25C
Q2  15,700 J
QTotal  Q1  Q2  50,100J 15,700J  65,800J 66,000J 66kJ
time (min)
T (°C)


2. Suppose in the Icy Hot lab that the burner transfers 325 kJ of energy to 450 g of liquid water at
 would be boiled away?
20.˚C. What mass of the water
2) some H2O boils
100
50
1) H2O warms to 100°C
0
tim e (m in)

Q1  m  c  T
Q1  450g 4.18 J gC  80C
Q1  150kJ
Q2  QT  Q1
Q2  325kJ  150kJ
Q2  m  H v
Q
m 2
Hv
175,000J
m
 77.4g  77g
2260 J g
Q2  175kJ
3. A 12oz can of soft drink (assume m = 340 g) at 25˚C is placed in a freezer where the temperature is

– 12 ˚C. How much energy must be removed from the soft drink for it to reach this temperature?
T (°C)

25
1
2

Q1  m  c  T
3
Q1  340 g  4.18 J
-25
tim e (m in)
3 steps:
1) cool H2O to 0°C
2) freeze H2O
3) cool ice
liquid :

©Modeling Instruction – AMTA 2012
freezing H 2O :
Q2  m  H f
 25C
gC
Q1  35,5300 J or 35.5 kJ
solid :
Q2  340 g  334 J g
Q2 113,560 J or 113.6 kJ
Q3  m  c  T
 35.5 kJ  113 .6 kJ  8.6 kJ
 158 .6 kJ or 160 kJ
Q3  340 g  2.1J /gC 12C

Q3  8568 J or 8.6 kJ
1
QTotal  Q1  Q2  Q3
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4. 65.0 kilojoules of energy are added to 150 g of ice at 0.0˚C. What is the final temperature of the
water?
T (°C)
30
? °C
2
10
1
final temperature :
melting ice :
Q1  m  H f
Q2  m  c  T
Q1  150 g  334 J / g
T 
Q2  QTotal  Q1
Q2

mc
14,900 J
T 
150 g  4.18 J / g C
Q2  65.0 kJ  50.1 kJ  14.9 kJ
T  23.8C
Q1  50.1 k J
-10
tim e (m in)
Q2  14,900 J
5. 250 kJ of energy are removed from a 4.00 x 102 g sample of water at 60˚C. Will the sample of
water completely freeze? Explain.
maximum mass :
Q2  m  H f
Q1  m  c  T
Q1  400 g  4.18 J / g C  60C
T (°C)
65
40
1
m
Q1  100 kJ
15
2
Q2  QTotal  Q1
-10
Q2  250 kJ  100 kJ  150 kJ
tim e (m in)
Q2 150,000 J

 449 g
Hf
334 J / g
Our mass is less than the maximum
mass 250kJ could freeze so we
could freeze even more than the
400g sample.
Q2  150,000 J
6. An ice cube tray full of ice (235g) at –7.0˚C is allowed to warm up to room temperature (22˚C).
1
How much energy must be absorbed by the contents of the tray in order for this to happen?
Q1  m  c  T
melting ice :
Q2  m  H f
Q1  235 g  2.10 J / g C  7C
Q 2  235 g  334 J / g
Q1  3.5 kJ
Q 2  78.5 k J
liquid :
Q3  m  c  T
QTotal  Q1  Q2  Q3
T (°C)
solid :
15
3
2
1
-10
tim e (m in)
 3.5 kJ  78.5 kJ  21.6 kJ
 104 kJ
Q3  235 g  4.18 J / g C  22C
Q3  21.6 kJ
If this same quantity of energy were removed from 40.0 g of water vapor at 100˚C, what would
be the final temperature of the water?
T (°C)
7.
125
100
75
50
25
0
1
2
? °C
tim e (m in)
condensing vapor:
Q1  m  H v
change in temperature :
Q2  m  c  T
Q1  40.0 g  2260 J / g
T 
Q1  90.4 kJ
Q2
13,600 J

m  c 40.0 g  4.18 J / g C
T  81.3C
Q2  QTotal  Q1
Q2  104 kJ  90.4 kJ  13.6 kJ
T final  100C  81.3C  18.7C
Q2  13,600 J
©Modeling Instruction – AMTA 2012
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WS 4 Quantitative Energy 2 Key