**Unit Four Homework Solutions, September 23. 2010**
**1 A piston-cylinder device whose piston is resting on top of a set of stops initially contains 0.5 kg of helium gas at 100 kPa and 25**^{o}C. The mass of the piston is such that 500 kPa of pressure is required to raise it. How much heat must be added to the helium before the piston starts rising.
In this problem, we are given the helium mass, m = 0.5 kg, which we assume is constant. We are also given the initial pressure and temperature, P_{1} = 100 kPa and T_{1} = 25^{o}C = 298.15 K. The fact that the piston is resting on stops as heat is added means that the volume is constant. Because we stop adding heat when the pressure is just enough to raise the piston, we know that the final pressure, P_{2} = 500 kPa, the pressure required to raise the piston. We also know that V_{2} = V_{1} because of the constant volume process.

We will assume that helium is an ideal gas at these low pressures. We can find the heat from the first law: Q = U + W = mc_{v}dT + W for an ideal gas. In this constant-volume process the work, W = 0. From Figure 4-24 on page 177, we see that the heat capacity of helium is essentially constant over a wide range of temperatures. Thus we can use the constant heat capacity of helium from Table A-2 on page 909: c_{v} = 3.1156 kJ/(kg∙K). The same table gives the gas constant for helium; R = 2.0769 kJ/(kg∙K).

From the equation that Q = mc_{v}dT + W, we see that Q = mc_{v}(T_{2} – T_{1}) in this case of constant heat capacity and no work. We do not know the final temperature, T_{2}, but we can find it from the ideal gas law (and V_{2} = V_{1}).

We can now find the heat transfer.

=

**1,857 kJ**.

**2 A mass of 15 kg of air in a piston-cylinder device is heated from 25 to 77**^{o}C by passing current through a resistance heater inside the cylinder. The pressure inside the cylinder is held constant at 300 kPa during the process and a heat loss of 60 kJ occurs. Determine the electrical energy supply in kWh.
The first law for this problem can be written as Q = U + W = U + W_{e,out} + PdV, where W_{e,out} is the electrical work done by the system. We expect that this will be negative since we see tghat there is an electrical energy input to the system. We can say that W_{e,in} = -W_{e,out} = U + PdV – Q.

For this constant pressure process, PdV = P_{1-2} (V_{2} – V_{1}) where P_{1-2} = P_{1} = P_{2} is the constant pressure of 300 kPa. With this work term we can write our first law as follows.

W_{e,in} = U + P_{1-2} (V_{2} – V_{1}) – Q = U_{2} – U_{1} + P_{2}V_{2} – P_{1}V_{1} – Q = H_{2} – H_{1} – Q = m(h_{2} – h_{1}) – Q

We can find the specific enthalpy change, h_{2} – h_{1} = c_{p}dT, where c_{p} can be found from Table A-2(b) on page 910. At the average temperature of (25^{o}C + 77^{o}C)/2 = 51^{o}C = 324.15 K, we can interpolate to find c_{p} = 1.0065 kJ/(kg∙K). With this value of c_{p} (and using Q = –60 kJ because it is a heat loss) we can find the electrical work input as follows.

**W**_{e,in}** = 0.235 kWh**.

**3 A room is heated by a baseboard resistance heater. When the heat losses from the room on a winter day amount to 6,500 kJ/h, the air temperature in the room remains constant, even though the heater operates continuously. Determine the power rating of the heater in kW.**

Since the temperature of the air is constant, the internal energy is constant. If we assume that the volume of the house does not change, there is no work. (Note that we also have to assume that this is a closed system; *i.e*. there is no air flow into or out of the room.) The first law then gives Q =U + W = 0. Thus the heat loss through the wall is balanced by the heat input from the electric heater. This input must be 6500 kJ/h = (6500 kJ/h)[(1 kW∙h) / (3600 kJ) ] = **1.81 kW**.

**4 A piston-cylinder device contains 3 ft**^{3} of air at 60 psia and 150^{o}F. Heat is transferred to the air in the amount of 40 Btu as the air expands isothermally. Determine the amount of boundary work during this process.
From the first law, Q = U + W, with U = mc_{v}dT for ideal gases. We see that U for this process is zero, since the temperature is constant. Thus, we have **W = Q = 40 Btu**.

**5**

**Problem 4-73**

**5**

A piston-cylinder device, with a set of stops on the top, initially contains 3 kg of air at 200 kPa and 27^{o}C. Heat is now transferred to the air, and the piston rises until it hits the stops, at which point the volume is twice the initial volume. More heat is transferred until the pressure inside the cylinder also doubles. Determine the work done and the amount of heat transfer for this process. Also, show the process on a P-v diagram.
From the problem description, we see that the path is a two step path. The first part is constant pressure as the piston rises to the stops. During this part, the volume doubles. The second part occurs at constant volume after the piston reaches the stops. (The actual data points plotted on the diagram are found in the calculations below.).

If we denote the initial point as P_{1} = 200 kPa and T_{1} = 27^{o}C = 300.15 K, and the final point as P_{2} = 2 P_{1} = 400 kPa and V_{2} = 2 V_{1}, the intermediate point, where the piston reaches the stops has V = V_{2} and P = P_{1}. The mass m = 3 kg.

All the work is done during the constant pressure part of the path. This work is simply found as the rectangular area under the path: W = P_{1}(V_{2} – V_{1}). The heat transfer is then found as Q = U + W = mu + P_{1}(V_{2} – V_{1}).

If we assume that air behaves as an ideal gas at these pressures, we can compute the initial volume from the ideal gas law.

Since we are given V_{2} = 2 V_{1}, we know that V_{2} = 2.584 m^{3}. We can now find the work as follows.

W = P_{1}(V_{2} – V_{1}) = (200 kPa)( 2.584 m^{3} – 1.292 m^{3}) = 258.4 kPa∙m^{3} = 258.4 kJ

To compute the change in internal energy, we need the final temperature. We can find this final temperature from the ideal gas law.

For this large temperature difference we should consider the temperature dependence of the heat capacity. We can use the ideal gas tables for air on pages 934-35. Interpolating in these tables we find the following values for the ideal gas internal energy:

u_{1} = u(300.15 K) = 214.18 kJ/kg u_{2} = u(1200.6 K) = 933.86 kJ/kg

We can now find the heat transfer, Q = U + W = m(u_{2} – u_{1}) + W = (3 kg)( 933.86 kJ/kg – 214.18 kJ/kg) + 258.4 kJ. This gives **Q = 2,417.5 kJ**.