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Q
4.2. Q = mc ΔT _
Q = mc ΔT ∴ ΔT =

mc
cal
gC°)
(
= (221 g) 0.093 _ (38.0°C – 20.0°C) = 0.20 kcal
__

(
)
kcal

_ (15.53 kg) 0.200 _
cal

= (221)(0.093)(18.0) g × × C° kgC°
gC°
_ _ _
0.20
__ kcal 1 kgC°

g·cal·C°

_ = × ×

= 370 (15.53)(0.200) 1 kg kcal
gC°
kcal·kgC°
_

= 370 cal = 0.0644
kcal·kg
4.3. First, you need to know the energy of the moving bike and = 6.4 × 10 C°
–2
rider. Since the speed is given as 36.0 km/h, convert to m/s by

multiplying by 0.2778 m/s per km/h: 4.5. The Calorie used by dietitians is a kilocalorie; thus, 250.0 Cal is

250.0 kcal. The mechanical energy equivalent is 1 kcal = 4,184 J,
_
m/s
( 36.0 0.2778 _ so (250.0 kcal)(4,184 J/kcal) = 1,046,250 J.
)(
km
)

h km/h
Since W = Fd and the force needed is equal to the weight (mg)
_ _ _ 2
km
m
h

= (36.0)(0.2778) × × of the person, W = mgh = 75.0 kg × 9.8 m/s ×
h km s
10.0 m = 7,350 J for each stairway climb.
= 10.0 m/s
A total of 1,046,250 J of energy from the French fries would
Th en require (1,046,250 J) ÷ (7,350 J per climb), or 142 trips up the
stairs.
1 _ 2
KE = mv

2 4.6. For unit consistency,
1 _ 2 5 _ 5 _ 5 _

= (100.0 kg)(10.0 m/s) T C = (T F – 32°) = (68° – 32°) = (36°) = 20°C

2 9 9 9
1 _ 2 2 5 _ 5 _
= (100.0 kg)(100 m /s ) = (32° – 32°) = (0°) = 0°C

2 9 9
kg·m
1 _ _ 2

= (100.0)(100) Glass bowl:

2 s 2
Q = mc ΔT
= 5,000 J
kcal
)
(

= (0.5 kg) 0.2 _ (20C°)
Second, this energy is converted to the calorie heat unit kgC°
through the mechanical equivalent of heat relationship, that kg
_ _ _
C°
1.0 kcal = 4,184 J, or that 1.0 cal = 4.184 J. Th us, = (0.5)(0.2)(20) × kcal ×

1 kgC° 1
5,000 J
_ = 2 kcal

4,184 J/kcal
Iron pan:
J
_ _
kcal

1.195 ×

1 J Q = mc ΔT
)
kcal
(
1.20 kcal = (0.5 kg) 0.11 _ (20C°)

kgC°
4.4. First, you need to find the energy of the falling bag. Since the _

kcal

potential energy lost equals the kinetic energy gained, the = (0.5)(0.11)(20) kg × × C°
kgC°
energy of the bag just as it hits the ground can be found from
= 1 kcal
PE = mgh

4.7. Note that a specific heat expressed in cal/gC° has the same
2
= (15.53 kg)(9.8 m/s )(5.50 m)
numerical value as a specific heat expressed in kcal/kgC°

kg·m because you can cancel the k units. You could convert
_

= (15.53)(9.8)(5.50) × m
s 2 896 cal to 0.896 kcal, but one of the two conversion methods is
needed for consistency with other units in the problem.
= 837 J
_
Q
In calories, this energy is equivalent to Q = mc ΔT ∴ m =
c ΔT
_
837 J
896 cal
= 0.20 kcal

= __

4,184 J/kcal _
( 0.056 gC°)
cal
(80.0C°)
Second, the temperature change can be calculated from the
gC°
1
equation giving the relationship among a quantity of heat (Q), __ cal _ _ _
896
= × ×

mass (m), specific heat of the substance (c), and the change of (0.056)(80.0) 1 cal C°
temperature: = 200 g
= 0.20 kg
654 APPENDIX E Solutions for Group A Parallel Exercises E-12