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Phys Int CC Ch 10 - Thermal Energy - Answers PDF

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CK-12 Physics Concepts - Intermediate
Answer Key
Chapter 10: Thermal Energy
10.1 Heat, Temperature, and Thermal Energy Transfer
Practice
Questions
1. Which material was a better conductor of heat?
2. Explain why metals feel cold even when they are at room temperature?
Answers
1. The aluminum melted the ice cube faster because it was able to conduct the heat
better.
2. The heat from your finger is able to be dissipated throughout the metal so more heat
leaves your body. When you touch styrofoam, the heat remains on the surface and it
feels warmer.
Review
Questions
1.
2.
3.
4.
Convert 4.22 K to °C.
Convert 37°C to K.
If you had beeswax attached to one end of a metal skewer and you placed the
other end of the skewer in a flame, what would happen after a few minutes?
Which contains more heat, a coffee cup of boiling water or a bathtub of room
temperature water?
Answers
1. 273.15 - 4.22 = 268.93 °C
2. 37 + 273.15 = 310.15 K
3. After a few minutes, the beeswax would begin to melt because of the heat
transfer along the metal skewer.
4. The bathtub has more heat, but a lower temperature.
10.2 Specific Heat
Practice
1
Added an interactive link
Review
Questions
1. How much heat is absorbed by 60.0 g of copper when it is heated from 20.0°C to
80.0°C?
2. A 40.0 kg block of lead is heated from -25°C to 200.°C. How much heat is
absorbed by the lead block?
3. The cooling system of an automobile motor contains 20.0 kg of water. What is
the
of the water if the engine operates until 836,000 J of heat have been
added to the water?
Answers
𝐽
1. Using 𝑄 = 𝑚𝑐∆𝑡 ∶ (0.06𝑘𝑔) (385 𝑘𝑔 𝐾) (60°) = 1,386 J
𝐽
2. Using 𝑄 = 𝑚𝑐∆𝑡 ∶ (40𝑘𝑔) (130 𝑘𝑔 𝐾) (225°) = 1,170,000 J
𝑄
3. Using ∆𝑡 = 𝑚𝑐 ∶
836,000𝐽
𝐽
𝐾)
𝑘𝑔
(20.0𝑘)(4180
∶
10.3 Calorimetry
Practice
Questions
1. What is the number 4.18 J/gºC in the video?
2. In the equation, q = mc∆t, what does c represent?
3. What does it mean if the temperature in the calorimeter goes down?
Answers
1. The number is water's specific heat capacity.
2. C represents the material's specific heat capacity.
3. Heat was absorbed and the reaction is endothermic.
Review
Questions
1.
A 300.0 g sample of water at 80.0ºC is mixed with 300.0 g of water at
10.0ºC. Assuming no heat loss to the surroundings, what is the final temperature
of the mixture?
2
2.
3.
A 400.0 g sample of methanol at 16.0ºC is mixed with 400.0 g of water at
85.0ºC. Assuming no heat loss to the surroundings, what is the final temperature
of the mixture? The specific heat of methanol is 2450 J/kg•ºC.
A 100.0 g brass block at 100.0ºC is placed in 200.0 g of water at 20.0ºC. The
specific heat of brass is 376 J/kg•ºC. Assuming no heat loss to the surroundings,
what is the final temperature of the mixture?
Answers
1.
2.
3.
The mixture contains equal volumes of the same material (water), so the
temperature will be the arithmetic mean of the two: 45 C.
Using 𝑚𝑚 𝑐𝑚 (𝑡2 − 𝑡1 )𝑚 = 𝑚𝑤 𝑐𝑤 (𝑡2 − 𝑡1 )𝑤 ∶
𝐽
𝐽
. 4𝑘𝑔 (2450 𝑘𝑔 𝐶) (16 − 𝑥) = .4𝑘𝑔 (4180 𝑘𝑔 𝐶) (𝑥 − 85) = 59.5 C.
Using 𝑚𝑏 𝑐𝑏 (𝑡2 − 𝑡1 )𝑏 = 𝑚𝑤 𝑐𝑤 (𝑡2 − 𝑡1 )𝑤 ∶
𝐽
𝐽
. 1𝑘𝑔 (376 𝑘𝑔 𝐶) (100 − 𝑥) = .2𝑘𝑔 (4180 𝑘𝑔 𝐶) (𝑥 − 20) = 23.4
C.
10.4 Change of State
Practice
Questions
1. For water, which takes more energy, melting or evaporating?
2. When are there two phases present at the same time in the pot?
Answers
1. Evaporation has a higher energy required.
2. At 0ºC and 100ºC, the water is in transition and two phases are present.
Review
Questions
1. A 200. g sample of water at 60.0°C is heated to water vapor at 140.0°C. How
much heat was absorbed?
2. A 175 g lump of molten lead at its melting point (327°C) is placed into 55.0 g of
water at 20.0°C. The specific heat of lead is 130 J/kg °C and the Hf of lead is
20,400 J/kg.
a. When the lead has become solid but is still at the melting point, what is the
temperature of the water?
b. When the lead and the water have reached equilibrium, what is the
temperature of the mixture?
3
Answers
𝐽
𝐽
1. Using (. 2𝑘𝑔 (4187 𝑘𝑔 𝐶) (100 − 60)) + (. 2𝑘𝑔 (2260000 𝑘𝑔)) +
𝐽
(. 2𝑘𝑔 (1996 𝑘𝑔 𝐶) (140 − 100)) = 5.01 x 105 J
2.
𝑄
a. Using 𝑚𝑐 = ∆𝑡 ∶
0.175𝑘𝑔∗20,400
𝐽
𝑘𝑔
𝐽
𝐶)
𝑘𝑔
.055𝑘𝑔(4187
= ∆𝑡 = 15.5 ∶ 20.0 + 15.5 = The water is
35.5 °C.
b. Using 𝑚𝑙 𝑐𝑙 (𝑡2 − 𝑡1 )𝑙 = 𝑚𝑤 𝑐𝑤 (𝑡2 − 𝑡1 )𝑤 ∶
𝐽
𝐽
.175𝑘𝑔 (130 𝑘𝑔 𝐶) (327 − 𝑥) = .055𝑘𝑔 (4180 𝑘𝑔 𝐶) (𝑥 − 35.5) At
equilibrium, the temperature is 61.8 °C.
4
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