Temperature and Phase Changes
Answers
Part A
1.
a)
How many kilojoules are required to heat 0.200 kg of Au from 15.0°C to 85.0°C?
Q = mc∆T
Q = (0.200 kg)(0.129 kJ/kg°C)(70.0°C) = 1.81 kJ
b) If the same amount of heat is added to 0.200 kg of water at 15.0°C, what will be the final temperature?
Q = mc∆T
Q
1.81 kJ
ΔT =
=
= 2.16°C
mc (0.200 kg )(4.18 kJ kg°C )
final T = 15.0°C + 2.16°C = 17.2°C
2.
How much heat is required to change the temperature of each of the following from 15.0°C to 60.0°C?
a)
35.0 g of water
Q = mc∆T
Q = (35.0 g)(4.18 J/g°C)(45.0°C) = 6583.5 J= 6580 K or 6.58 kJ
b) 35.0 g of Pyrex glass (cpyrex = 0.840 kJ/kg•°C)
Q = mc∆T
Q = (35.0 g)(0.840 J/g°C)(45.0°C) = 1323 J = 1320 K or 1.32 kJ
c)
35.0 g of Pt
Q = mc∆T
Q = (35.0 g)(0.133 J/g°C)(45.0°C) = 209.475 J = 209 J or 0.209 kJ
3.
Find the energy needed to heat 150.0 g of water from 25.3°C to 75.0°C.
Q = mc∆T
Q = (150.0 g)(4.18 J/g°C)(49.7°C) = 31 161.9 J = 31 200 J or 31.2 kJ
Chemistry 12 – L. Farrell - Thermochemistry – Temperature and Phase Changes – Answers – Page 1 of 10
4.
2
5.00 x 10 g of ice were cooled from –10.0°C to –41.5°C. Find the heat released.
Q = mc∆T
Q = (500. g)(2.06 J/g°C)(-31.5°C) = -32 445 J
Q = -32 400 J or –32.4 kJ
or 32.4 kJ released
5.
50.0 g of steam at 140.0°C were cooled, releasing 2.50 kJ of heat. Find the new temperature of the steam.
Q = mc∆T
Q
- 2.50 kJ
ΔT =
=
= −24.8°C
mc (0.0500 kg )(2.02 kJ kg°C )
final T = 140.0°C − 24.8°C = 115.2°C
6.
250.0 g of pure gold was heated from 21.8°C to 55.3°C. Find the heat energy required.
Q = mc∆T
Q = (250.0 g)(0.129 J/g°C)(33.5°C) = 1080.375 J = 1080 J = 1.08
7.
a)
kJ
How many kilojoules are required to heat 0.100 kg of Cu from 10.0°C to 100.0°C?
Q = mc∆T
Q = (0.100 kg)(0.385 kJ/kg°C)(90.0°C) = 3.46 kJ
b) The same quantity of heat is added to 0.100 kg of Al at 10.0°C. Which metal reaches the higher temperature,
the Cu or the Al?
The copper reaches the higher temperature because it has the lower specific
heat capacity. The specific heat capacity is equal to the amount of energy
required to change the temperature of one gram of a substance by one
degree Celsius, therefore, the smaller the heat capacity, the larger the
temperature change for the same amount of added heat.
8.
1.50 kJ of heat energy was applied to 10.0 g of copper at 30.0°C. The same amount of energy was used to heat
10.0 g of aluminum at 30.0°C. Which metal reaches the highest temperature?
The copper reaches the higher temperature because it has the lower specific
heat capacity. The specific heat capacity is equal to the amount of energy
required to change the temperature of one gram of a substance by one degree
Celsius, therefore, the smaller the heat capacity, the larger the temperature
change for the same amount of added heat.
Chemistry 12 – L. Farrell - Thermochemistry – Temperature and Phase Changes – Answers – Page 2 of 10
Part B
9.
Find the heat energy needed to melt 48.0 g of ice at 0.00°C.
n=
mass
48.0 g
=
= 2.66 mol
molar mass 18.01528 g mol
ΔH = nHfus = (2.66 mol)(6.03 kJ mol) = 16.1 kJ
10. How much heat energy is released when 1.00 kg of steam at 100.0°C is condensed to water at 100.0°C?
mass
1000 g
=
= 55.5 mol
molar mass 18.01528 g mol
q = nHvap = (55.5 mol)(- 40.8 kJ mol) = −2260 kJ
n=
q = - 2260 kJ
or 2260 kJ released
2
11. Find the heat energy needed to melt 4.00 x 10 g of copper at its melting point. The molar heat of fusion for
copper is 13.0 kJ/mol.
n=
mass
400. g
=
= 6.29 mol
molar mass 63.546 g mol
q = nHfus = (6.29 mol)(13.0 kJ mol) = 81.8 kJ
12. What mass of ethanol could be vapourized at its boiling point of 78.3°C if 250.0 kJ of heat energy was consumed?
The molar heat of vapourization of ethanol is 40.5 kJ/mol.
q = nHfus
n=
q
250.0 kJ
=
= 6.17 mol
Hfus 40.5 kJ mol
mass = (mol)(molar mass ) = (6.17 mol)(46.06904 g mol) = 284 g C 2H5OH
Chemistry 12 – L. Farrell - Thermochemistry – Temperature and Phase Changes – Answers – Page 3 of 10
13. Find the molar heat of fusion of sodium chloride if 103.3 kJ of heat energy was required to melt 200.0 g of sodium
chloride at its melting point of 802.0°C.
n=
mass
200.0 g
=
= 3.42 mol
molar mass 58.44277 g mol
q = nHfus
q 103.3 kJ
Hfus = =
= 30.2 kJ mol
n 3.42 mol
Part C
14. Calculate the amount of energy to melt 5.00 kg of aluminum pop cans initially at 25.0°C. The molar enthalpy of
fusion of aluminum is 10.8 kJ/mol.
q = (25.0°C 660.°C) + (melt)
q = mc∆T + nHfus
q = (5.00 kg)(0.900 kJ/kg°C)(635°C) + (185.3 mol)(10.8 kJ/mol)
q = 2857 kJ + 2001.4 kJ
q = 4860 kJ
15. How much heat is given up when 20.0 g of steam at 120.0°C is cooled to –15.0°C?
q = (120.0°C 100.0°C) + (condense) + (100.0°C 0.0°C) + (freeze) + (0.0°C -15.0°C)
q = mc∆T + nHvap + mc∆T + nHfus + mc∆T
q = (0.0200 kg)(2.02 kJ/kg°C)( –20.0°C) + (1.11 mol)( –40.8 kJ/mol)
+ (0.0200 kg)(4.18 kJ/kg°C)( –100.0°C) + (1.11 mol)( –6.03 kJ/mol)
+ (0.0200 kg)(2.06 kJ/kg°C)( –15.0°C)
q = –0.808 kJ – 45.29 kJ – 8.36 kJ – 6.69 kJ – 0.618 kJ
q = -61.8 kJ
or 61.8 kJ released
Chemistry 12 – L. Farrell - Thermochemistry – Temperature and Phase Changes – Answers – Page 4 of 10
16. How much heat is required to convert 45.0 g of ice at –115.0°C into steam at 220.0°C?
q = (-115.0°C 0.0°C) + (melt) + (0.0°C 100.0°C) + (boil) + (100.0°C 220.0°C)
q = mc∆T + nHfus + mc∆T + nHvap + mc∆T
q = (0.0450 kg)(2.06 kJ/kg°C)(115.0°C) + (2.50 mol)(6.03 kJ/mol)
+ (0.0450 kg)(4.18 kJ/kg°C)(100.0°C) + (2.50 mol)(40.8 kJ/mol)
+ (0.0450 kg)(2.02 kJ/kg°C)(120.0°C)
q = 10.6605 kJ + 15.0622 kJ + 18.81 kJ + 101.913 kJ + 10.908 kJ
q = 157 kJ
17. Find the heat required when 80.0 g of water at 25.0°C are converted to steam at 100.0°C.
q = (0.0°C 100.0°C) + (boil)
q = mc∆T + nHvap
q = (0.0800 kg)(4.18 kJ/kg°C)(75.0°C) + (4.44 mol)(40.8 kJ/mol)
q = 25.08 kJ + 181.18 kJ
q = 206 kJ
18. a)
How many kilojoules are required to heat 0.450 kg of Au from –25.0°C to 215.0°C?
q = mc∆T
q = (0.450 kg)(0.129 kJ/kg°C)(240.0°C)
q = 13.9 kJ
b) If the same amount of heat is added to 0.0350 kg of ice at –10.0°C, what will be the final water temperature?
q = (-10.0°C 0.0°C) + (melt) + (0.0°C T)
q = mc∆T + nHfus + mc∆T
13.9 kJ = (0.0350 kg)(2.06 kJ/kg°C)(10.0°C) + (1.94 mol)(6.03 kJ/mol)
+ (0.0350 kg)(4.18 kJ/kg°C)(T – 0.0°C)
13.9 kJ = 0.721 kJ + 11.71 kJ + 0.1463 kJ/°C T – 0 kJ
1.498 kJ = 0.1463 kJ/°C T
T = 10.2°C
Chemistry 12 – L. Farrell - Thermochemistry – Temperature and Phase Changes – Answers – Page 5 of 10
19. Find the heat required to heat 40.0 g of ice at –25.0°C and convert it to steam at 130.0°C
q = (-25.0°C 0.0°C) + (melt) + (0.0°C 100.0°C) + (boil) + (100.0°C 130.0°C)
q = mc∆T + nHfus + mc∆T + nHvap + mc∆T
q = (0.0400 kg)(2.06 kJ/kg°C)(25.0°C) + (2.22 mol)(6.03 kJ/mol)
+ (0.0400 kg)(4.18 kJ/kg°C)(100.0°C) + (2.22 mol)(40.8 kJ/mol)
+ (0.0400 kg)(2.02 kJ/kg°C)(30.0°C)
q = 2.06 kJ + 13.39 kJ + 16.72 kJ + 90.59 kJ + 2.42 kJ
q = 125 kJ
20. 36.0 g of steam at 100.0°C was converted to water at 25.0°C. Find the heat released.
q = (condense) + (100.0°C 25.0°C)
q = nHvap + mc∆T
q = (1.998 mol)( –40.8 kJ/mol) + (0.0360 kg)(4.18 kJ/kg°C)( –75.0°C)
q = –81.53 kJ – 11.29 kJ
q = -92.8 kJ
or
92.8 kJ released
Part D
21. 1.00 g of anthracite coal gives off about 30.6 kJ when burned. What mass of coal is required to heat 3.50 L of
water from 10.0°C to 95.0°C?
heat gained by water = heat lost by coal
water (10°C 95°C)
q = mc∆T
q = (3.50 kg)(4.18 kJ/kg°C)(85.0°C)
q = 1243.55 kJ
∴the coal must supply 1240 kJ
1 g coal
− 30.6 kJ
=
x
- 1243.55 kJ
x = 40.6 g of coal needed
Chemistry 12 – L. Farrell - Thermochemistry – Temperature and Phase Changes – Answers – Page 6 of 10
22. Determine the resulting temperature when 150.0 g of ice at –20.0°C is mixed with 9.00 kg of water at 50.0°C.
heat gained by the ice = heat lost by the water
(-20.0°C 0.0°C) + (melt) + (0.0°C T) = (50.0°C T)
q = –q
mc∆T + nHfus + mc∆T = -[mc∆T]
(0.150 kg)(2.06 kJ/kg°C)(20.0°C) + (8.32 mol)(6.03 kJ/mol)
+ (0.150 kg)(4.18 kJ/kg°C)(T – 0.0°C) = –[(9.00 kg)(4.18 kJ/kg°C)(T - 50.0°C)]
6.18 kJ + 50.2 kJ + 0.627 kJ/°C T – 0 kJ = –[37.62 kJ/°C T – 1881 kJ]
6.18 kJ + 50.2 kJ + 0.627 kJ/°C T – 0 kJ = 1881 kJ – 37.62 kJ/°C T
37.62 kJ/°C T + 0.627 kJ/°C T = 1881 kJ – 6.18 kJ – 50.2 kJ
38.247 kJ/°C T = 1824.62 kJ
T = 47.7°C
23. Determine the resulting temperature when 1.00 kg of ice at –20.0°C is mixed with 3.60 kg of water at 65.0°C.
heat gained by the ice = heat lost by the water
(-20.0°C 0.0°C) + (melt) + (0.0°C T) = (65.0°C T)
q = –q
mc∆T + nHfus + mc∆T = –[mc∆T]
(1.00 kg)(2.06 kJ/kg°C)(20.0°C) + (55.5 mol)(6.03 kJ/mol)
+ (1.00 kg)(4.18 kJ/kg°C)(T – 0.0°C) = –[(3.60 kg)(4.18 kJ/kg°C)(T – 65.0°C)]
41.2 kJ + 334.7 kJ + 4.18 kJ/°C T – 0 kJ = –[15.05 kJ/°C T – 978.12 kJ]
41.2 kJ + 334.7 kJ + 4.18 kJ/°C T – 0 kJ = 978.12 kJ – 15.05 kJ/°C T
15.05 kJ/°C T + 4.18 kJ/°C T = 978.12 kJ – 41.2 kJ – 334.7 kJ
19.228 kJ/°C T = 602.22 kJ
T = 31.3°C
Chemistry 12 – L. Farrell - Thermochemistry – Temperature and Phase Changes – Answers – Page 7 of 10
24. a)
If 150.0 g of water at 0.00°C is added to 100.0 g of water at 90.0°C, what will be the final temperature of the
water?
(0.0°C T) = (90.0°C T)
q = –q
mc∆T = –[mc∆T]
(150.0 g)(4.18 J/g°C)(T – 0.0°C) = –[(100.0 g)(4.18 J/g°C)(T – 90.0°C)]
627 J/°C T – 0 J = –[418 J/°C T – 37620 J]
627 J/°C T = 37620 J - 418 J/°C T
627 J/°C T + 418 J/°C T = 37620 J
1045 J/°C T = 37620 J
T = 36.0°C
b) If 250.0 g of Au at 100.0 °C is added to 500.0 g of water at 5.00°C, what will be the final temperature of the
Au?
(100.0°C T) = (5.00°C T)
q = –q
mc∆T = –[mc∆T]
(250.0 g)(0.129 J/g°C)(T – 100.0°C) = –[(500.0 g)(4.18 J/g°C)(T – 5.00°C)]
32.25 J/°C T – 3225 J = –[2090 J/°C T – 10450 J]
32.25 J/°C T – 3225 J = 10450 J - 2090 J/°C T
32.25 J/°C T + 2090 J/°C T = 10459 J + 3225 J
2122.25 J/°C T = 13675 J
T = 6.44°C
Chemistry 12 – L. Farrell - Thermochemistry – Temperature and Phase Changes – Answers – Page 8 of 10
25. Coal provides 30.5 kJ of energy per gram burned. Find the mass of coal required to heat 5.00 L of water from
10.0°C to 85.0°C.
heat gained by water = heat lost by coal
water (10°C 85°C)
q = mc∆T
q = (5.00 kg)(4.18 kJ/kg°C)(75.0°C)
q = 1567.5 kJ
∴the coal must supply 1567.5 kJ
1 g coal − 30.6 kJ
=
x
- 1567.5 kJ
x = 51.4 g of coal needed
26. 5.00 L of water at 70.0°C was cooled by adding 0.500 kg of ice at –10.0°C. Calculate the final temperature.
(-10.0°C 0.0°C) + (melt) + (0.0°C T) = (70.0°C T)
q = –q
mc∆T + nHfus + mc∆T = –[mc∆T]
(0.500 kg)(2.06 kJ/kg°C)(10.0°C) + (27.75 mol)(6.03 kJ/mol)
+ (0.500 kg)(4.18 kJ/kg°C)(T – 0.0°C) = –[(5.00 kg)(4.18 kJ/kg°C)(T – 70.0°C)]
10.3 kJ + 167.36 kJ + 2.09 kJ/°C T – 0 kJ = –[20.9 kJ/°C T – 1463 kJ]
10.3 kJ + 167.36 kJ + 2.09 kJ/°C T = 1463 kJ – 20.9 kJ/°C T
2.09 kJ/°C T + 20.9 kJ/°C T = 1463 kJ – 10.3 kJ – 167.36 kJ
22.99 kJ/°C T = 1285.34 kJ
T = 55.9°C
Chemistry 12 – L. Farrell - Thermochemistry – Temperature and Phase Changes – Answers – Page 9 of 10
27. 250.0 g of water at 32.5°C had the following metals placed in it:
78.0 g of silver at 90.0°C
39.5 g of aluminum at 78.0°C
55.0 g of copper at 85.0°C
Calculate the final temperature of all substances.
Heat gained by water = heat lost by silver + aluminum + copper
(32.5°C T) = (90.0°C T) + (78.0°C T) + (85.0°C T)
–q = q
–[mcΔT] = mc∆T + mc∆T + mc∆T
–[(250.0 g)(4.18 J/g°C)(T – 32.5°C)] = (78.0 g)(0.237 J/g°C)(T – 90.0°C)
+ (39.5 g)(0.900 J/g°C)(T – 78.0°C) + (55.0 g)(0.385 J/g°C)(T – 85.0°C)
–[1045 J/°C T – 33962.5 J] = 18.486 J/°C T – 1663.74 J
+ 35.55 J/°C T – 2772.9 J + 21.175 J/°C T – 1799.875 J
–1045 J/°C T + 33 962.5 J = 75.211 J/°C T – 6236.515 J
33 962 J + 6236.515 J = 75.211 J/°C T + 1045 J/°C T
40 199.015 J = 1120.211 J/°C T
T = 35.9°C
28. Water at 50.0°C was cooled to 0.00°C by adding 18.02 g of ice at 0.00°C. Calculate the mass of water cooled.
(Assume the ice just melts.)
29. When 1.00 g of propane is burned, about 2.36 kJ of heat is given off. What mass of water at 50.0°C can be
converted into super-heated steam at 170.0°C when 4.00 mol of propane are burned?
30. If 50.0 g of steam at 165.0°C is added to 200.0 g of ice at –45.0°C, what will be the final temperature of the
system?
31. If 85.0 g of steam at 130.0°C is added to 300.0 g of ice at –20.0 °C, what will be the final temperature of the
system?
32. The average body temperature of a healthy human is 37.0°C. What mass of steam at 140.0°C must be added to
250.0 g of ice at –30.0°C to produce water having the same temperature as a human body?
33. What mass of water at 50.0°C was cooled, and converted to ice at –20.0°C, if 84.7 kJ of heat energy was released?
34. What mass of steam at 100.0°C is required to heat 400.0 g of water from 25.0°C to 60.0°C?
35. What mass of ice at –10.0°C must be added to 125 g of steam at 220.0°C to produce water at 100.0°C?
36. Calculate the mass of steam at 180.0°C that is required to raise the temperature of 0.470 kg of V from –5.00°C to
60.0°C?
Chemistry 12 – L. Farrell - Thermochemistry – Temperature and Phase Changes – Answers – Page 10 of 10