1 Btu 0.252 kcal 1055 J.

Heat Problems
1.
How much heat (in joules) is required to raise the temperature of 30.0 kg of water
from 15°C to 95°C?
2.
To what temperature will 7700 J of heat raise 3.0 kg of water that is initially at
10.0°C?
3.
An average active person consumes about 2500 Cal a day. (a) What is this in joules?
(b) What is this in kilowatt-hours? (c) Your power company charges about a dime
per kilowatt-hour. How much would your energy cost per day if you bought it from
the power company? Could you feed yourself on this much money per day?
4.
A British thermal unit (Btu) is a unit of heat in the British system of units. One Btu
is defined as the heat needed to raise 1 lb of water by 1 F°. Show that
1 Btu = 0.252 kcal = 1055 J.
5.
A water heater can generate 32, 000 kJ h. How much water can it heat from 15°C to
50°C per hour?
6.
A small immersion heater is rated at 350 W. Estimate how long it will take to heat a
cup of soup (assume this is 250 mL of water) from 20°C to 60°C.
7.
How many kilocalories are generated when the brakes are used to bring a 1200-kg
car to rest from a speed of 95 km h?
Specific Heat; Calorimetry
8.
An automobile cooling system holds 16 L of water. How much heat does it absorb if
its temperature rises from 20°C to 90°C?
9.
What is the specific heat of a metal substance if 135 kJ of heat is needed to raise 5.1
kg of the metal from 18.0°C to 31.5°C?
10. Samples of copper, aluminum, and water experience the same temperature rise when
they absorb the same amount of heat. What is the ratio of their masses? [Hint: See
Table 14–1.]
11. A 35-g glass thermometer reads 21.6°C before it is placed in 135 mL of water.
When the water and thermometer come to equilibrium, the thermometer reads
39.2°C. What was the original temperature of the water?
12. What will be the equilibrium temperature when a 245-g block of copper at 285°C is
placed in a 145-g aluminum calorimeter cup containing 825 g of water at 12.0°C?
13. A hot iron horseshoe (mass = 0.40 kg ), just forged (Fig. 14–16), is dropped into
1.35 L of water in a 0.30-kg iron pot initially at 20.0°C. If the final equilibrium
temperature is 25.0°C, estimate the initial temperature of the hot horseshoe.
14. A 215-g sample of a substance is heated to 330°C and then plunged into a 105-g
aluminum calorimeter cup containing 165 g of water and a 17-g glass thermometer
at 12.5°C. The final temperature is 35.0°C. What is the specific heat of the
substance? (Assume no water boils away.)
15. How long does it take a 750-W coffeepot to bring to a boil 0.75 L of water initially
at 8.0°C? Assume that the part of the pot which is heated with the water is made of
360 g of aluminum, and that no water boils away.
16.
Estimate the Calorie content of 75 g of candy from the following measurements.
A 15-g sample of the candy is allowed to dry before putting it in a bomb calorimeter. The
aluminum bomb has a mass of 0.725 kg and is placed in 2.00 kg of water contained in an
aluminum calorimeter cup of mass 0.624 kg. The initial temperature of the mixture is
15.0°C, and its temperature after ignition is 53.5°C.
17. When a 290-g piece of iron at 180°C is placed in a 95-g aluminum calorimeter cup
containing 250 g of glycerin at 10°C, the final temperature is observed to be 38°C.
Estimate the specific heat of glycerin.
18. The 1.20-kg head of a hammer has a speed of 6.5 m s just before it strikes a nail
(Fig. 14–17) and is brought to rest. Estimate the temperature rise of a 14-g iron nail
generated by 10 such hammer blows done in quick succession. Assume the nail
absorbs all the energy.
19. A 0.095-kg aluminium sphere is dropped from the roof of a 45-m-high building. If
65% of the thermal energy produced when it hits the ground is absorbed by the
sphere, what is its temperature increase?
20. The heat capacity, C, of an object is defined as the amount of heat needed to raise its
temperature by 1 C°. Thus, to raise the temperature by ΔT requires heat Q given by
Q = C ΔT .
(a) Write the heat capacity C in terms of the specific heat, c, of the material. (b)
What is the heat capacity of 1.0 kg of water? (c) Of 25 kg of water?
Latent Heat
21. How much heat is needed to melt 16.50 kg of silver that is initially at 20°C?
22. During exercise, a person may give off 180 kcal of heat in 30 min by evaporation of
water from the skin. How much water has been lost?
23. If 2.80 × 10 5 J of energy is supplied to a flask of liquid oxygen at −183º C, how much
oxygen can evaporate?
24. A 30-g ice cube at its melting point is dropped into an insulated container of liquid
nitrogen. How much nitrogen evaporates if it is at its boiling point of 77 K and has a
latent heat of vaporization of 200 kJ kg? Assume for simplicity that the specific heat
of ice is a constant and is equal to its value near its melting point.
25. A cube of ice is taken from the freezer at −8.5º C and placed in a 95-g aluminum
calorimeter filled with 310 g of water at room temperature of 20.0°C. The final
situation is observed to be all water at 17.0°C. What was the mass of the ice cube?
26. An iron boiler of mass 230 kg contains 830 kg of water at 18°C. A heater supplies
energy at the rate of 52,000 kJ h. How long does it take for the water (a) to reach the
boiling point, and (b) to all have changed to steam?
27. In a hot day’s race, a bicyclist consumes 8.0 L of water over the span of four hours.
Making the approximation that all of the cyclist’s energy goes into evaporating this
water as sweat, how much energy in kcal did the rider use during the ride? (Since
the efficiency of the rider is only about 20%, most of the energy consumed does go
to heat, so our approximation is not far off.)
28. What mass of steam at 100°C must be added to 1.00 kg of ice at 0°C to yield liquid
water at 20°C?
29. The specific heat of mercury is 138 J kg ⋅ Cº . Determine the latent heat of fusion of
mercury using the following calorimeter data: 1.00 kg of solid Hg at its melting
point of −39.0º C is placed in a 0.620-kg aluminum calorimeter with 0.400 kg of
water at 12.80°C; the resulting equilibrium temperature is 5.06°C.
30.
A 70-g bullet traveling at 250 m s penetrates a block of ice at 0°C and comes to rest
within the ice. Assuming that the temperature of the bullet doesn’t change
appreciably, how much ice is melted as a result of the collision?
31. A 54.0-kg ice-skater moving at 6.4 m s glides to a stop. Assuming the ice is at 0°C
and that 50% of the heat generated by friction is absorbed by the ice, how much ice
melts?
32. At a crime scene, the forensic investigator notes that the 8.2-g lead bullet that was
stopped in a doorframe apparently melted completely on impact. Assuming the
bullet was fired at room temperature (20°C), what does the investigator calculate as
the minimum muzzle velocity of the gun?
Solutions to Problems
1.
The kcal is the heat needed to raise 1 kg of water by 1 Co. Use the definition to find the heat needed.
1 kcal " 4186 J %
= 1.0 × 10 7 J
( 30.0 kg ) ( 95º C − 15º C)
(1 kg ) (1 Cº ) $# 1 kcal '&
2.
The kcal is the heat needed to raise 1 kg of water by 1 Co. Use that definition to find the temperature
change. Then the final temperature can be found.
! 7700 J $ ! 1 kcal $ (1 kg ) (1 Cº )
= 0.61Cº → Final Temperature = 10.6ºC
#" 3.0 kg &% #" 4186 J &% 1 kcal
3.
(a)
(b)
" 4.186 × 10 3 J %
7
2500 Cal $
'& = 1.0 × 10 J
1 Cal
#
! 1 kWh $
2500 Cal #
= 2.9 kWh
" 860 Cal &%
(c)
4.
5.
6.
7.
At 10 cents per day, the food energy costs $0.29 per day . It would be practically impossible to
feed yourself in the United States on this amount of money.
Assume that we are at the surface of the Earth so that 1 kg has a weight of 2.20 lb.
1 kcal
! 0.454 kg $ ! 5 9º C $
1 Btu = (1 lb ) (1º F ) #
= 0.252 kcal
" 1 lb &% #" 1º F &% (1 kg ) (1 Cº )
! 4186 J $
0.252 kcal #
= 1055 J
" 1 kcal &%
The energy input is causing a certain rise in temperature, expressible as a number of Joules per hour
per Co. Convert that to mass using the definition of kcal.
" 3.2 × 10 7 J h % " 1 kcal % (1 kg ) (1 Cº )
= 2.2 × 10 2 kg h
$#
'& $# 4186 J '& 1 kcal
35 Cº
The wattage rating is Joules per second. Note that 1 L of water has a mass of 1 kg.
)
, 1 kcal # 4186 J & # 1 s &
# 1 kg &
−1
2
+( 2.50 × 10 L ) %$ 1 L (' ( 40Cº ) . (1 kg ) (1 Cº ) %$ kcal (' %$ 350 J (' = 1.2 × 10 s = 2.0 min
*
The energy generated by using the brakes must equal the car’s initial kinetic energy, since its final
kinetic energy is 0.
2
(
" 1 m s % + " 1 kcal %
2
1.2 × 10 3 kg *( 95 km h ) $
'& - $# 4186 J '& = 1.0 × 10 kcal
3.6
km
h
#
)
,
The heat absorbed can be calculated from Eq. 14-2. Note that 1 L of water has a mass of 1 kg.
*
$ 1 × 10 −3 m 3 ' $ 1.0 × 10 3 kg ' 6
Q = mcΔT = ,(16 L ) &
3
)( &%
)( / ( 4186 J kg • Cº ) ( 90º C − 20º C ) = 4.7 × 10 J
1
L
1
m
%
+
.
The specific heat can be calculated from Eq. 14-2.
Q
1.35 × 10 5 J
Q = mcΔT → c =
=
= 1961 J kg • Cº ≈ 2.0 × 10 3 J kg • C º
mΔT ( 5.1 kg ) ( 31.5º C − 18.0º C )
The heat absorbed by all three substances is given by Eq. 14-2, Q = mcΔT . Thus the amount of mass
Q = 12 mv02 =
8.
9.
10.
1
2
(
)
Q
. The heat and temperature change are the same for all three substances.
cΔT
Q
Q
Q
1 1
1
1
1
1
: mAl : mH2 O =
:
:
=
:
:
=
:
:
cCu ΔT cAl ΔT cH2 O ΔT cCu cAl cH2 O 390 900 4186
can be found as m =
mCu
4186 4186 4186
:
:
= 10.7 : 4.65 :1
390 900 4186
The heat gained by the glass thermometer must be equal to the heat lost by the water.
mglass cglass Teq − Ti glass = mH2 O cH2 O TiH2 O − Teq
=
11.
(
( 35 g ) ( 0.20
)
(
)
(
cal g • Cº ) ( 39.2º C − 21.6º C ) = (135 g ) (1.00 cal g • Cº ) TiH2 O − 39.2º C
TiH2 O = 40.1º C
)
12.
The heat lost by the copper must be equal to the heat gained by the aluminum and the water.
mCu cCu TiCu − Teq = mAl cAl Teq − Ti Al + mH2 O cH2 O Teq − Ti H2 O
(
)
( 0.245 kg ) ( 390
(
(
)
)
"( 0.145 kg ) ( 900 J kg • Cº ) %
J kg • Cº ) 285º C − Teq = $
' Teq − 12.0º C
$# + ( 0.825 kg ) ( 4186 J kg • Cº ) '&
(
)
(
)
Teq = 19.1º C
13.
The heat lost by the horseshoe must be equal to the heat gained by the iron pot and the water. Note
that 1 L of water has a mass of 1 kg.
mshoe cFe Tishoe − Teq = mpot cFe Teq − Ti pot + mH2 O cH2 O Teq − TiH2 O
(
)
( 0.40 kg ) ( 450
(
(
)
)
J kg • Cº ) (Ti shoe − 25.0º C ) = ( 0.30 kg ) ( 450 J kg • Cº ) ( 25.0Cº − 20.0Cº )
+ (1.35 kg ) ( 4186 J kg • Cº ) ( 25.0Cº − 20.0Cº )
Ti shoe = 186º C ≈ 190º C
14.
The heat lost by the substance must be equal to the heat gained by the aluminum, water, and glass.
mx cx Ti x − Teq = mAl cAl Teq − Ti Al + mH2 O cH2 O Teq − Ti H2 O + mglass cglass Teq − Ti glass
(
cx =
)
(
(
)
mAl cAl Teq − Ti Al + mH2 O
(
)
c (T
m (T
)
)
(
H2 O
eq
− Ti H2 O + mglass cglass Teq − Ti glass
x
ix
− Teq
)
(
)
)
"( 0.105 kg ) ( 900 J kg • Cº ) + ( 0.165 kg ) ( 4186 J kg • Cº ) + ( 0.017 kg ) ( 840 J kg • Cº ) $% ( 22.5 Cº )
=#
( 0.215 kg ) ( 330º C − 35.0º C)
= 2.84 × 10 2 J kg • Cº
15.
The heat must warm both the water and the pot to 100oC. The heat is also the power times the time.
Q = Pt = mAl cAl + mH2 O cH2 O ΔTH2 O →
(
t=
(m
)
)
c + mH2 O cH2 O ΔTH2 O
Al Al
P
#( 0.36 kg ) ( 900 J kg • Cº ) + ( 0.75 kg ) ( 4186 J kg • Cº ) %& ( 92Cº )
=$
750 W
= 425 s ≈ 7 min
16.
The heat released by the 15 grams of candy in the burning is equal to the heat absorbed by the bomb,
calorimeter, and water.
Q15 = !" mbomb + mcup cAl + mH2 O cH2 O #$ ΔT
(
)
= !"( 0.725 kg + 0.624 kg ) ( 0.22 kcal kg • Cº ) + ( 2.00 kg ) (1.00 kcal kg • Cº ) #$ ( 53.5º C − 15.0º C )
= 88.43 kcal
The heat released by 75 grams of the candy would be 5 times that released by the 15 grams.
Q75 = 5Q15 = 5(88.43 kcal)= 440 kcal = 440 Cal
17.
The heat lost by the iron must be the heat gained by the aluminum and the glycerin.
mFe cFe Ti Fe − Teq = mAl cAl Teq − Ti Al + mgly cgly Teq − Ti gly
(
(
)
( 0.290 kg ) ( 450
)
(
)
J kg • Cº ) (142Cº ) = ( 0.095 kg ) ( 900 J kg • Cº ) ( 28Cº ) + ( 0.250 kg ) cgly ( 28Cº )
3
cgly = 2.3 × 10 J kg • C º
18.
We assume that all of the kinetic energy of the hammer goes into heating the nail.
2
KE = Q → 10 ( 12 mhammer vhammer
) = mnailcFe ΔT →
ΔT =
19.
10
(
1
2
2
mhammer vhammer
)=
2
5 (1.20 kg ) ( 6.5 m s )
= 40.24 Cº ≈ 4.0 × 101 Cº
( 0.014 kg ) ( 450 J kg • Cº )
mnail cFe
65% of the original potential energy of the aluminum goes to heating the aluminum.
0.65PE = Q → 0.65mAl gh = mAl cAl ΔT →
ΔT =
(
)
2
0.65gh 0.65 9.80 m s ( 45 m )
=
= 0.32 Cº
cAl
( 900 J kg • Cº )
20.
(a)
Since Q = mcΔT and Q = CΔT , equate these two expressions for Q and solve for C.
Q = mcΔT = CΔT →
21.
C = mc
(b)
For 1.0 kg of water:
C = mc = (1.0 kg )(4186 J kg • Cº )= 4.2 × 10 3 J C º
(c)
For 25 kg of water:
C = mc = (25 kg )(4186 J kg • Cº )= 1.0 × 10 5 J C º
The silver must be heated to the melting temperature and then melted.
Q = Qheat + Qmelt = mcΔT + mLfusion
(
)
= (16.50 kg ) ( 230 J kg • Cº ) ( 961º C − 20º C ) + (16.50 kg ) 0.88 × 10 5 J kg = 5.0 × 10 6 J
22.
Assume that the heat from the person is only used to evaporate the water. Also, we use the heat of
vaporization at room temperature (585 kcal/kg), since the person’s temperature is closer to room
temperature than 100oC.
Q
180 kcal
Q = mLvap → m =
=
= 0.308 kg ≈ 0.31 kg = 310 mL
Lvap 585 kcal kg
23.
The oxygen is all at the boiling point, so any heat added will cause oxygen to evaporate (as opposed
to raising its temperature). We assume that all the heat goes to the oxygen, and none to the flask.
Q
2.80 × 10 5 J
=
= 1.3 kg
Lvap 2.1× 10 5 J kg
Assume that all of the heat lost by the ice cube in cooling to the temperature of the liquid nitrogen is
used to boil the nitrogen, and so none is used to raise the temperature of the nitrogen. The boiling
point of the nitrogen is 77 K = −196º C
mice cice Ti ice − T f ice = mnitrogen Lvap →
Q = mLvap → m =
24.
(
)
(
mice cice Ti ice − T f
mnitrogen =
25.
Lvap
) = ( 3.0 × 10
−2
)
kg ( 2100 J kg • Cº ) ( 0º C − −196º C )
200 × 10 3 J kg
= 6.2 × 10 −2 kg
The heat lost by the aluminum and 310 g of liquid water must be equal to the heat gained by the ice in
warming in the solid state, melting, and warming in the liquid state.
mAl cAl Ti Al − Teq + mH2 O cH2 O Ti H2 O − Teq = mice "# cice (Tmelt − Ti ice ) + Lfusion + cH2 O Teq − Tmelt $%
( 0.095 kg ) ( 900 J kg • Cº ) ( 3.0 Cº ) + ( 0.31 kg ) ( 4186 J kg • Cº ) ( 3.0 Cº )
mice =
= 9.90 × 10 −3 kg
5
#"( 2100 J kg • Cº ) ( 8.5Cº ) + 3.3 × 10 J kg + ( 4186 J kg • Cº ) (17 Cº ) %$
(
26.
ice
(a)
(
)
)
(
)
The heater must heat both the boiler and the water at the same time.
Q1 = Pt1 = mFe cFe + mH2 O cH2 O ΔT →
(
t1 =
(m
)
)
c + mH2 O cH2 O ΔT
Fe Fe
P
#( 230kg ) ( 450 J kg • Cº ) + ( 830 kg ) ( 4186 J kg • Cº ) %& ( 82 Cº )
=$
5.2 × 10 7 J h
= 5.642 h ≈ 5.6 h
(b)
Assume that after the water starts to boil, all the heat energy goes into boiling the water, and
none goes to raising the temperature of the iron or the steam.
mH2 O Lvap ( 830 kg ) ( 22.6 × 10 5 J kg )
Q2 = Pt 2 = mH2 O Lvap → t 2 =
=
= 36.073 h
P
5.2 × 10 7 J h
Thus the total time is t1 + t 2 = 5.642 h + 36.073 h = 41.72 h ≈ 42 h
27.
We assume that the cyclist’s energy is only going to evaporation, not any heating. Then the energy
needed is equal to the mass of the water times the latent heat of vaporization for water. Note that 1 L
of water has a mass of 1 kg. Also, we use the heat of vaporization at room temperature (585 kcal/kg),
since the cyclist’s temperature is closer to room temperature than 100oC.
Q = mH 2O Lvap = (8.0 kg )(585 kcal kg )= 4.7 × 10 3 kcal
28.
The heat lost by the steam condensing and then cooling to 20oC must be equal to the heat gained by
the ice melting and then warming to 20oC.
msteam "# Lvap + cH2 O Ti steam − Teq $% = mice "# Lfus + cH2 O Teq − Ti ice $%
(
)
(
(
(
)
)
5
" Lfus + cH O Teq − Ti ice $
2
% = 1.00 kg "# 3.33 × 10 J kg + ( 4186 J kg • Cº ) ( 20Cº ) $%
msteam = mice #
(
)
" Lvap + cH O Ti steam − Teq $
"# 22.6 × 10 5 J kg + ( 4186 J kg • Cº ) ( 80 Cº ) $%
2
#
%
)
= 1.61 × 10 −1 kg
29.
The heat lost by the aluminum and the water must equal the heat needed to melt the mercury and to
warm the mercury to the equilibrium temperature.
mAl cAl Ti Al − Teq + mH2 O cH2 O Ti H2 O − Teq = mHg "# Lfusion + cHg Teq − Tmelt $%
(
Lfusion =
(
)
(
)
(
(
)
mAl cAl Ti Al − Teq + mH2 O cH2 O Ti H2 O − Teq
) − c (T
Hg
mHg
eq
− Tmelt
)
)
"( 0.620 kg ) ( 900 J kg • Cº ) + ( 0.400 kg ) ( 4186 J kg • C º ) $% (12.80º C − 5.06º C )
=#
1.00 kg
− (138 J kg • C º ) "# 5.06º C − ( −39.0º C ) $%
= 1.12 × 10 4 J kg
30.
Assume that the kinetic energy of the bullet was all converted into heat which melted the ice.
2
1
→
2 mbullet v = Q = mice Lfusion
31.
1
2
(
2
)
)
2
mskater v 2 14 ( 54.0 kg ) ( 6.4 m s )
=
= 1.7 × 10 −3 kg = 1.7 g
Lfusion
3.33 × 10 5 J kg
The kinetic energy of the bullet is assumed to warm the bullet and melt it.
2
1
→
2 mv = Q = mcPb (Tmelt − Ti ) + mLfusion
mice =
32.
(
−2
1
mbullet v 2 2 7.0 × 10 kg ( 250 m s )
=
= 6.6 × 10 −3 kg = 6.6 g
Lfusion
3.33 × 10 5 J kg
Assume that all of the melted ice stays at 0oC, so that all the heat is used in melting ice, and none in
warming water. The available heat is half of the original kinetic energy
2
1 1
= Q = mice Lfusion →
2 2 mskater v
mice =
1
4
(
)
v = 2 #$ cPb (Tmelt − Ti ) + Lfusion %& = 2 #$(130 J kg • Cº ) ( 327º C − 20º C ) + 0.25 × 10 5 J kg %&
2
= 3.6 × 10 m s