Solutions

1)

(a)

$\Delta L = \alpha L \Delta T=0.0044 ~\rm cm$.

(b)

Since $A=L^2$, $A+\Delta A = (L+\Delta L)^2$ where $\Delta L = \alpha L \Delta T$, you can express $\Delta A \approx 2L^2\alpha \Delta T$. So

$${\Delta A/A \over \Delta T}= 2\alpha~.$$

(c)

Some air bubbles get trapped inside the liquid after you shake the bottle. If you suddenly open the bottle, the air expands due to sudden drop of the pressure (at a much higher rate than the liquid), which pushes some of the liquid out of the bottle.

2)

(a)

Equate the heat loss of the metal and the heat gain of the liquid:

$$c_l(T_f- T_l)m_l = c_x(T_x-T_f)m_x~,$$

then the specific heat of the metal $c_x$ is

$$c_x ={c_l(T_f-T_l)m_l\over (T_x-T_f)m_x}~.$$

(b)

You only need the info on melting of silver. With specific heat $c_s$ and the specific heat of fusion $L_f$, the total heat needed to melt the silve of mass $m$ at temperature $T$ is

$$Q= m[c_s(T_m-T)+L_f]~,$$

where $T_m$ is the melting point. Then $Q=4.27\times 10^4 ~\rm J$.

(c)

In constant volume processes, no work is done by the gas, and therefore, $\Delta E_{\rm int}=Q$. Then $C_V = dE_{\rm int}/dT = 3RT$ per mol.

3)

(a)

In one cycle, $\Delta E_{\rm int}=Q-W=0$, which means that the total net heat added is equal to the total net work done by the gas. Since the cycle is counterclockwise, the net work is negative (negative heat), $W=\oint PdV = -{1\over 2}3\times 20 =-30 ~\rm J=Q$.

(b) to (d):

The internal energy difference in this case is $\Delta E_{\rm int} ={3\over 2} (P_fV_f - P_iV_i)=Q-W$, and $W=\int PdV$, from which you can find the heat $Q$. (b) An isobaric process: $\Delta E_{\rm int}= {3\over 2}P \Delta V$ and $W= P\Delta V= - 90~\rm J$. So $Q={5\over 2 }P\Delta V = -225~\rm J$. (c) An isovolumetric process: $\Delta E_{\rm int} ={3\over 2}V\Delta P$, and $W=0$. Then $Q= \Delta E_{\rm int} = -30~\rm J$. (d) $\Delta E_{\rm int} ={3\over 2} (P_fV_f - P_iV_i)=165 ~\rm J$, and $W=\int PdV = 70 ~\rm J$. Then $Q= W+\Delta E_{\rm int} = 235 ~\rm J$.

4)

(a)

The mass of one mole of the air is $M=28.9\times 10^{-3}~\rm kg/mol$. The specific heat is then $C_V/M=719 ~\rm J/kg\cdot ^o C$.

(b)

From the equipartition theorem, $C_V = N_fR/2$ where $N_f$ is the number of degrees of freedom of each molecule. So $N_f = 5$ in this case. For the kinetic energy only, however, the average value is still the same as that of an ideal monatomic gas, ${1\over 2}m\bar{v^2} = {3\over 2}k_B T$. Therefore, $\sqrt{\bar{v^2}} = \sqrt{3k_BT\over m}=\sqrt{3RT/M}=29.4 ~\rm m/s$.

(c)

Since the weight of the piston would not change over this process, this is an isobaric process, with the well-known relation $C_P=C_V+R$. The air inside the cylinder is $n=PV/RT = 28.1 ~\rm mol$, The total heat needed to be added is $Q=C_P\Delta T n = 3.27\times 10^5 ~\rm J$.

5)

(a)

Since $e=1-\vert Q_C/Q_H\vert$, you only need to calculate the total added heat and the total extracted heat. It is helpful to draw a diagram and number each stroke. The processes 1 and 2 adds heat to the engine, 3 and 4 extracts heat. ($Q_H=Q_1+Q_2$, $Q_C=Q_3+Q_4$.)

Since 1 is an isovolumetric process and 2 is an isothermal process, $Q_1 = nC_V\Delta T = 3nRT_1$, and $Q_2=W=\int PdV =3nRT_1\ln (4)$. So $Q_H = 3nRT_1 [1+\ln (4) ]$.

Similarly, 3 is an isovolumetric process and 4 is an isothermal process. So $Q_3 = 3nT_1nR$, $Q_4 =nRT_1\ln(4)$, and finally $Q_C= nRT_1[3+\ln(4)]$.

$$e={2\over 3} {\ln(4)\over 1+\ln (4)}~.$$

(b)

The total entropy change of a reversible cycle is zero. In realistic engines with irreversible processes, the entropy change of the total system is always positive.