CHAPTER 10 SOLUTIONS TO PROBLEMS AND QUESTIONS

Citation: Jacqueline D. Spears and Dean Zollman, Instructor's Guide for The Fascination of Physics, (The Benjamin /Cummings Publishing Company, Inc., Menlo Park, CA 1985). Permission granted by the publisher and authors.

A. Review of Chapter Material

A1. The answers to the "A1" questions are round directly in the text. Look for each term printed in boldface type.

A2. A change Or state occurs when thermal energy is added to or removed from a material at its melting or boiling point.

A3. A temperature change occurs when thermal energy is added to or removed from a material which is not undergoing a change of state.

A4. Temperature scales differ in the size of the degree and in the location or the origin (O) of the scale.

A5. The specific heat capacity of the sauce is much higher than that of the crust. Thus when both the sauce and the crust are at the same temperature, the sauce will have stored much more thermal energy than has the crust. When you touch each of them, the sauce will transfer much more thermal energy than the crust.

A6. Thermal energy always moves spontaneously from the warmer substance to the cooler one.

A7. When a substance is not undergoing a change in state, the amount of thermal energy transferred depends on the substance's mass, specific heat capacity, and change in temperature.

A8. Thermal energy is needed to separate the molecules of the substance.

A9. When a substance changes state, the amount of thermal energy transferred depends on the substance's mass and latent heat.

A10. We know that thermal energy has been transferred when we see either a change in the temperature of the substance or a change in its state.

B. Using the Chapter Material

B1. At -20º C the kinetic energy of the molecules is greater than it is for the same substance at -50º C.

B2. The molecules found in steam are not bound to one another whereas molecules in water are bound together, giving water its definite volume.

B3. The difference between the boiling temperature and the freezing temperature is 100 degrees. Thus, the size of the degree on this scale is the same as on the Celsius scale. The origin of the two scales is different.

B4. The specific heat capacity of the apples is more than three times greater than that of the aluminum. (See Table 10-1.) Because both materials have the same mass and temperature, this difference in specific heat capacity determines the difference in the thermal energy stored in them. More thermal energy is stored in the apples than in the aluminum pan. The apples are more likely to burn your hand.

B5. Because both materials have identical masses and temperatures, the difference in thermal energy stored in them is due to the difference in their specific heat capacities. From Table 10-1 we see that wood has a higher specific heat capacity than brick. So, it would provide more thermal energy.

B6. Thermal energy transferred = mass x specific heat capacity x temperature change = 70 kg x (3.3 kJ/º C kg) x (2º C) = 462 kJ.

B7. The final temperature (30º C) is halfway between the temperature of the air in the car and the temperature of the air from the air conditioner. Thus, one-half of the air (0.5 kg) must be replaced.

B8. At 100º C, steam contains much more energy than water at the same temperature. The latent heat of vaporization has been added to each kilogram of water at 100º C to convert it to steam at the same temperature.

B9. Energy transferred = mass x (latent heat of vaporization) = (0.3 kg) x (156 kJ/kg) = 46.8 kJ .

B10. When water freezes, thermal energy is transferred from the water to the air in the freezer. Thus, the temperature of the air in the freezer must increase. When ice melts, energy must be transferred to the ice from the surrounding substance. Thus, the temperature of the air must decrease.

B11. Thermal energy transferred = mass x specific heat capacity x temperature change = 70 kg x (3.3 kJ/º C kg) x temperature change = 6,000 kJ. Solving for temperature change we find that the temperature change = 26º c. It's a good thing that we sweat a lot.

C. Extensions to New Situations

C1. (a) The energy saved is equal to the energy needed to heat the excess 20 kg of water. Thermal energy = (mass) x (specific heat) x (change in temperature) = 20 kg x (4.2 kJ/º C kg) x (40º C - 20º C) = 1680 kJ. (b) Thermal energy = (mass) x (specific heat) x (change in temperature) = 165 kg x (4.2 kJ/º C kg) x (40º C - 20º C) = 13,860 kJ. This amount of energy will be transferred to the air in the bathroom, heating the house instead of the sewers.

C2. (a) Thermal energy = (mass) x (specific heat) x (change in temperature) = 0.4 kg x (1.0 kJ/º C kg) x (170º C - 20º C) = 60 kJ. (b) Thermal energy = (mass) x (specific heat) x (change in temperature) = 50 kg x (0.5 kJ/º C kg) x (170º C - 20 c) = 3750 kJ. (c) Thermal energy = (mass) x (specific heat) x (change in temperature) = 2 kg x (3.4 kJ/º C kg) x (170º C - 20º C) = 1020 kJ. (d) The total energy is the sum of the energies calculated above = 60 kJ + 3750 kJ + 1020 kJ = 4830 kJ. (e) The microwave oven would require only the energy to heat the rood (1020 kJ) and not the remaining 4830 kJ - 1020 kJ = 3810 kJ.

C3. (a) Thermal energy for air in conventional oven = (mass) x (specific heat) x (change in temperature) = 0.4 kg x (1.0 kJ/º C kg) x (170º C - 20º C) = 60 kJ. Thermal energy for air in countertop oven = (mass) x (specific heat) x (change in temperature) = 0.1 kg x (1.0 kJ/º C kg) x (170º C - 20º C) = 15 kJ. (b) The countertop oven will use less energy just because it contains less air to heat.

C4. (a) Thermal energy = latent heat of fusion x mass = 320 kJ/kg x 2 kg = 640 kJ. (b) Thermal energy = (mass) x (specific heat) x (change in temperature) - 2 kg x (4.2 kJ/º C kg) x (100º C - 0º C) = 840 kJ. (c) Thermal energy = latent heat of vaporization x mass = 2160 kJ/kg x 2 kg = 4320 kJ. (d) The most energy was needed to turn the water at 100º C into steam at 100º C. The least energy was needed to convert 0º C ice into 0º C water. (e) The total energy = 640 kJ + 840 kJ + 2160 kJ = 3640 kJ.

C5. (a) A typical beverage is water, so: Thermal energy removed from beverage = (mass) x (specific heat) x (change in temperature) = .35 kg x (4.2 kJ/º C kg) x (20º C - 5º C) = 22.05 kJ. (b) Thermal energy removed from aluminum can = (mass) x (specific heat) x (change in temperature) = 0.02 kg x (0.9 kJ/º C kg) x (20º C - 5º C) = 0.27 kJ. Thermal energy removed from steel can = (mass) x (specific heat) x (change in temperature) = 0.05 kg x (0.5 kJ/º C kg) x (20º C - 5º C) = 0.375 kJ. Thermal energy removed from glass = (mass) x (specific heat) x (change in temperature) = 0.20 kg x (0.8 kJ/º C kg) x (20º C - 5º C) = 2.4 kJ. (c) aluminum: 22.05 kJ + .27 kJ = 22.32 kJ; steel: 22.05 kJ + 0.375 kJ = 22.425 kJ ; glass: 22.05 kJ + 2.4 kJ = 24.45 kJ. (d) I would prefer aluminum cans because they require the least amount of thermal energy transferred.

C6. (a) As the ball drops, it gains kinetic energy. When it hits the ground and stops, that energy must be transformed into other forms of energy. Here, kinetic energy is transformed into the thermal energy of the snowball and of the ground. If the ball has sufficient kinetic energy, it will gain enough thermal energy to melt it. (b) Thermal energy needed = latent heat of fusion x mass = 320 kJ/kg x 2 kg = 640 kJ. (c) The gravitational potential energy will be the energy which is converted into kinetic energy then into thermal energy. So, by calculating the gravitational potential energy we can determine if at any of these heights the snowball has sufficient energy to melt when it reaches the ground. GPE = mass x acceleration due to gravity x height = 2 kg x 9.8 (m/s)/s X 4 m = 78.4 J (not enough); GPE = mass x acceleration due to gravity x height = 2 kg x 9.8 (m/s)/s X 400 m = 7840 J (still not enough); GPE = mass x acceleration due to gravity x height = 2 kg x 9.8 (m/s)/s X 40,000 m = 784,000 J. The kinetic energy that the snowball has just before impact will equal its gravitational potential energy before it was released. In order to melt upon impact, the snowball must have had 64 kJ or 640,000 joules of gravitational potential energy before being dropped. (d) The leftover energy would go into heating the water, so the melted snowball would have a temperature greater than 0º C.

C7. (a) At 4º C Freon-11 would be in a liquid state, while Freon-14 would be a gas. (b) If each entered as a liquid and were just at a temperature to turn into a gas, they would absorb energy = latent heat of vaporization x mass. Freon-11: thermal energy absorbed = 181 kJ/kg x 1 kg = 181 kJ ; Freon-14: thermal energy absorbed = 136 kJ/kg x 1 kg = 136 kJ. (c) The Freon-11 will make the better coolant in this case.

C8. The rubbing process involves transferring kinetic energy to the ice. Through the frictional interaction, this kinetic energy is converted into thermal energy. The thermal energy is transferred to the ice and causes it to melt.

C9. (a) As the weight drops, the gravitational potential energy of the weight is converted into kinetic energy of the paddles in the water. Through the frictional interaction between the water and the paddles, this energy of motion is converted into thermal energy. (b) Gravitational potential energy = mass x acceleration due to gravity x height = 26 kg x 9.8 (m/s)/s x 1.5 m = 382.2 J. (c) He applied the law of conservation of energy. The gravitational potential energy disappeared and the temperature of the water increased.