Chapter 12 Solutions. Q Reason: We ll use Equation Q = McΔT and solve for M. We are given. In each case we want to solve for.

Size: px

Start display at page:

Download "Chapter 12 Solutions. Q Reason: We ll use Equation Q = McΔT and solve for M. We are given. In each case we want to solve for."

Catherine Briggs
5 years ago
Views:

1 Chapter 12 Solutions Q Reason: Assume the gas is an ideal gas, and use the ideal gas law pv = nrt. Since the number of moles doesn t change and R is a constant, then Equation gives In each case we want to solve for (a) For and So the absolute temperature has increased by a factor of six. (b) For and So the absolute temperature has increased by a factor of 3/2. Assess: A shortcut is that if and we double V and triple p, then T is increased by a factor of 2 3 = 6. Q Reason: To measure the specific heat of a material by putting it in water, you need to know the mass of the sample, the mass of the water, the initial temperatures and the final temperature. The mass of the pennies was measured, but one was lost so that the mass of the sample was overestimated. When you solve the calorimetry equation, you obtain the following: Since you are using a value of which is too high, and since is in the denominator, your value of will be too low. Thus you will underestimate the specific heat. The answer is A. Assess: Another way to look at this is that since you are using too few pennies, the temperature of the water will rise less than it would have if you had used all of them. In a calorimetry problem, the greater the specific heat of a sample, the more influence it has on the final temperature. Since the sample will change the temperature of the water less than it would have with all the pennies, you will underestimate its specific heat. Q Reason: We ll use Equation Q = McΔT and solve for M. We are given Table 12.4 provides c = 4190 J/kg K. and The correct answer is A. Assess: For the water (with a large specific heat) to rise in temperature by that much in one minute the mass must be pretty small (72 g of water has a volume of about 1/3 of a cup). To boil a liter (just over a quart, and almost 14 times the answer to this question) of water on your stove in just a few minutes you need to deliver much more than 100 W to it.

2 Q Reason: Assume that the beakers are well-insulated. The specific heat of water is much higher than the specific heat of aluminum, so more heat will be taken from the beaker with the water added than the beaker with the aluminum added. In the beaker with ice added, the ice first melts and then the meltwater will have its temperature raised. Since the melting takes extra heat, the second beaker will end up with the lowest temperature. The correct choice is B. Assess: This is as expected from experience. Q Reason: The amount of heat needed can be calculated with Equation The heat the microwave oven provides is given by the power absorbed multiplied by the time it has been on, Q = PΔt. Setting this equal to the heat needed to raise the water s temperature and solving for the time, The correct choice is B. Assess: This makes sense, from common experience. Q Reason: Choice A can t be right because the question explicitly states both the steam and liquid water are at 100 C. But as the steam condenses on the skin it transfers a lot of heat to the skin; this causes the more severe burn. The correct answer is B. Assess: The specific heat of steam is actually less than that of liquid water. Problems: P12.2. Prepare: A water molecule has a single oxygen atom, while an oxygen molecule is made up of two oxygen atoms. A single mole of oxygen gas will contain two moles of oxygen atoms, since there are two oxygen atoms per molecule. There is one oxygen atom per molecule of water. In order to have the same number of oxygen atoms in the water as in one mole of gas, there must be two moles of water. A molecule of water consists of two hydrogen atoms and one oxygen atom and has a mass of Two moles of water has a mass of The mass of water required is 36.0 g. Assess: Note that since an oxygen molecule has a mass of 16 u + 16 u = 32 u, the mass of one mole of oxygen is a little less than the mass of water required. P12.4. Prepare: One liter is 1000 cubic centimeters. Assess: Note that the entire conversion factor must be cubed. P12.8. Prepare: We ll assume that air is an ideal gas so we can use the ideal gas law, We are given and Also recall that and oxygen makes up 20% of the air. Solve Equation for n, the number of moles of air.

3 Multiply the number of moles of air by 20% to get the number of moles of oxygen: (0.20 mol)(0.20) = mol of oxygen. Assess: The answer is a small number of moles of oxygen, but a large number of molecules of oxygen. P Prepare: Equation 12.9 gives the force due to a pressure applied over an area. The preliminary calculation is to compute the cross-section area of the tube. We are given Assess: 2.5 N is not a large force, but it is pushing on delicate tissue, so it pays to be careful. P Prepare: We need the mass of a mole of CO 2. Since carbon has an atomic mass of 12 and oxygen has an atomic mass of 16, the molecular mass of CO 2 is 44. Hence a mole of CO 2 has a mass of or We use Equation for rms speed but modify it by multiplying numerator and denominator by Avogadro s number. Assess: This is a typical speed for a gas molecule with a temperature in the hundreds of Kelvins. For example, the rms speed of an O 2 molecule at room temperature is about P Prepare: Assume the gas to be an ideal gas. Please refer to Figure P We will make use of the following conversions: 1 atm = Pa and 1 cm 3 = m 3. (a) Because the volume stays unchanged, the process is isochoric. (b) The ideal-gas law gives The final temperature T 2 is calculated as follows for an isochoric process: P Prepare: We ll compute the energy necessary to evaporate 3.5 L and then divide by an hour to get the rate. One liter of water has a mass of one kilogram. The rate is the energy divided by the time. Assess: That is an impressive power output, but necessary to keep cool in tropical climates. Notice that the value given for L f at body temperature is different from the one given in Table 12.5 for standard temperature (0 C).

4 P Prepare: We need to find the temperature increase and we are given the metabolic power the time of exercise and the man s mass We need the formula for power, which in this case is and Equation for heat absorbed. Here the heat produced by the exercise goes to increasing the thermal energy of the man, so we can write the following: For the specific heat, we use the mammalian specific heat from Table 12.4: First we will find the thermal energy produced by the exercise by solving the power equation. We can solve the equation for To two significant figures, his temperature increases by Assess: Even though he only exercised for 30 min, the man s temperature has increased by or This would bring a temperature of up to about which would be very dangerous. P Prepare: Heat will leave the patient and enter the ice, melting it. We need the formula for heat absorbed in melting and the formula for the heat lost by the patient For the specific heat, we will use the value for mammals from Table 12.4, The calorimetry equation is as follows: We need to solve for It takes of ice to reduce the fever by Assess: Since the ratio of the mammalian specific heat to the latent heat of fusion of ice is about it follows that a rule of thumb is that the amount of ice needed to cool a person by body mass. This is seen to be the case here. is one hundredth of their P Prepare: Treat the gas in the sealed container as an ideal gas and use Equation (a) From the ideal gas law equation pv = nrt, the volume V of the container is (b) The before-and-after relationship of an ideal gas in a sealed container (constant volume) is Assess: Note that gas-law calculations must use T in kelvins and pressure must be in Pa. P Prepare: We find the heat which must be supplied to the air from Equation for constant pressure processes. The value of is given in Table 12.6: The number of moles of air

5 inhaled can be obtained from the ideal gas law. We also find the final volume from the ideal gas law and subtract the initial volume to obtain (a) Solving the ideal gas law for the number of moles using the initial volume and temperature, we obtain the following: The amount of heat which must be supplied is (b) From the ideal gas law as stated in Equation 12.14, the final volume of the air is given by the following: s The increase in volume is Assess: We could have estimated the change in volume using Equation 12.18, but this would not have been as accurate since depends on temperature (it is given for in Table 12.3) and here the temperature varies from to P Prepare: The rate of heat transfer of solar energy is Q/Δt, which is equal to solar power. That is, the heat absorbed is (solar power) Δt which is also equal to Mc water ΔT according to Equation From the given information, we can then easily find Δt. The area of the garden pond is A = π(2.5 m) 2 = m 2 and the mass of water in the pond is kg. The water absorbs all the solar power, which is This power is used to raise the temperature of the water. That is, Assess: It is a common experience that a pool of water takes a few hours to warm up. A value of 8.7 h to heat 5900 kg water by 10 C is reasonable. P Prepare: Heat loss by conduction can be calculated with Equation Equation applies to heat lost by radiation. (a) The dead layer of air separates and insulates you from the air in the room. The conduction through the air layer is (b) The heat lost through radiation is given by Equation Body temperature is temperature of the walls is The (c) The heat lost to radiation is greater. (d) If the person is metabolizing food at a rate of he feels chilly because he is producing heat at a rate of and losing heat at a rate of or Assess: Putting some clothes on would decrease the heat lost by radiation, convection, and conduction.

Chapter 10 Temperature and Heat

Chapter 10 Temperature and Heat Thermodynamics deals with 1. Temperature. 2. The transfer and transformation of energy. 3. The relationship between macroscopic properties and microscopic dynamics. Temperature