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4.8. Since a watt is defined as a joule/s, finding the total energy in = 5,000 cal
joules will tell the time: = 5.00 kcal
Q = mc ΔT Q 2 = mL v
cal _
(
gC°)
g )
(
cal
= (250.0 g) 1.00 _ (60.0C°) = (250.0 g) 540.0

g·cal
_ _
cal
= (250.0)(1.00)(60.0) g × × C° = 250.0 × 540.0

gC° g
4
= 1.50 × 10 cal = 135,000 cal

This energy in joules is = 135.0 kcal
_
J
(
)
4

(1.50 × 10 cal) 4.184 = 62,800 J Q Total = Q 1 + Q 2
cal = 5.00 kcal + 135.0 kcal
_ J

A 300 watt heater uses energy at a rate of 300 , so = 140.0 kcal
s
62,800 J
_ = _ = 3.48 min, or
209 s

209 s is required, which is

s _
300 J/s 60 4.12. To change 20.0°C water to steam at 125.0°C requires three
min separate quantities of heat. First, the quantity Q 1 is the amount
1 _ of heat needed to warm the water from 20.0°C to 100.0°C (ΔT =

about 3 min 80.0°C). Th e quantity Q 2 is the amount of heat needed to take
2
Q
_ 100.0°C water to steam at 100.0°C. Finally, the quantity Q 3 is the
4.9. Q = mc ΔT ∴ c =
m ΔT amount of heat needed to warm the steam from 100.0°C to
__ 125.0°C. According to Table 4.4, the c for steam is 0.480 cal/gC°.
60.0 cal
=

(100.0 g)(20.0C°)
_
60.0
__ cal m = 100.0 g Q 1 = mc ΔT
=

(
cal
(100.0)(20.0) gC° ΔT water = 80.0C° _
(80.0C°)
= (100.0 g) 1.00 gC°)

_
cal
= 0.0300 ΔT steam = 25.0C°
gC° _
cal

L v(water) = 540.0 cal/g = (100.0)(1.00)(80.0) g × × C°
4.10. Since the problem specified a solid changing to a liquid without gC°

c steam = 0.480 cal/gC°
a temperature change, you should recognize that this is a _
g·cal·C°

= 8,000
question about a phase change only. The phase change from gC°

solid to liquid (or liquid to solid) is concerned with the latent
= 8,000 cal
heat of fusion. For water, the latent heat of fusion is given as
80.0 cal/g, and = 8.00 kcal
Q 2 = mL v
m = 250.0 g Q = mL f
(
cal _
g )

g )
L f (water) = 80.0 cal/g cal _ = (100.0 g) 540.0
(
= (250.0 g) 80.0
Q = ? _
g·cal
g·cal
_ = 100.0 × 540.0 g
= 250.0 × 80.0
g
= 54,000 cal
= 20,000 cal = 20.0 kcal
= 54.00 kcal
4.11. To change water at 80.0°C to steam at 100.0°C requires two
Q 3 = mc ΔT
separate quantities of heat that can be called Q 1 and Q 2 . Th e
(
gC°)
cal
quantity Q 1 is the amount of heat needed to warm the water = (100.0 g) 0.480 _ (25.0C°)

from 80.0°C to the boiling point, which is 100.0°C at sea level
cal

pressure (ΔT = 20.0C°). The relationship between the variables = (100.0)(0.480)(25.0) g × _ × C°

involved is Q 1 = mc ΔT. Th e quantity Q 2 is the amount of heat gC°
needed to take 100.0°C water through the phase change to _
g·cal·C°

steam (water vapor) at 100.0°C. The phase change from a liquid = 1,200 g·C°

to a gas (or gas to liquid) is concerned with the latent heat of
= 1,200 cal
vaporization. For water, the latent heat of vaporization is given
as 540.0 cal/g. = 1.20 kcal
m = 250.0 g Q 1 = mc ΔT Q total = Q 1 + Q 2 + Q 3
= 8.00 kcal + 54.00 kcal + 1.20 kcal
L v (water) = 540.0 cal/g
gC°)
(
cal
Q = ? = (250.0 g) 1.00 _ (20.0C°) = 63.20 kcal
_
cal
= (250.0)(1.00)(20.0) g × × C°

gC°
g·cal·C°
_

= 5,000
gC°
E-13 APPENDIX E Solutions for Group A Parallel Exercises 655