Homework 5

4-3

$L= 4200~\rm ft$ and $h=470~\rm ft$ at 50 F. The equation of the parabolic curve is $y(x) = {4h\over L^2} x^2$. The length of the cable is

$$S(h) = \int_{-L/2}^{L/2} dx~ \sqrt{1+( dy/dx)^2} ={L^2\over 8... ...nh^{-1}(4h/L) +{1\over 2} \sinh[ 2 \sinh^{-1}(4h/L)] \right\}~.$$

At 50 F, $S= 4292 ~\rm ft$, which is very close to the distance between the towers. This is because the sag is very small. (a) From $\Delta S \approx \alpha S\Delta T$, $\Delta S = 2.2~\rm ft$. (b) At $x=L/2$, the $xy$ slop of the cable is $dy(L/2)/dx \approx 4h/L =0.45$. Therefore, the increase in $h$ is approximately $\Delta h \approx 0.98~\rm ft.$

4-4

Molecular weight of water is $18 ~\rm g/mol$, so $9~\rm g$ is $0.5 ~\rm moles$. From $PV= nRT$, $P=1.61 \times 10^6 ~\rm Pa$.

4-5

From $P_1V_1/T_1 = P_2V_2/T_2$, $V_2 = V_1 ( P_1T_2/P_2 T_1)=1.51 \times 10^3~\rm m^3$, which means that $R_2 = 7.1 ~\rm m$.

4-6

From $PV= nRT$, we find that the mole number of the gas is $m=0.21 ~\rm mol$. (a) With the given initial pressure, volume, and temperature $P_0,V_0,T_0$, the spring constant $k$, and the cross-sectional area $A$, we set up an equation of state from the new pressure and the new volume formed by the increase in the height $h$:

$$(P_0+kh/A)(V_0+hA)=mRT$$

where $T=523~\rm K$. Solving this quadratic equation for $h$, we get

$$h= {1\over 2k}\left[ -(AP_0+kV_0/A)+\sqrt{ (AP_0+kV_0/A)^2+4kmR(T-T_0)} \right] =0.17~\rm m~.$$

(b) The new pressure is $P_0+kh/A = 1.35 \times 10^5 ~\rm Pa$.

5-1

Let the specific heat of the metal $c_m$, and that of water $c_w$. At the end of the day (at equilibrium), the heat gain of the water ( $M_w =14~\rm kg$) + container ( $M_c = 3.6~\rm kg$) is the same as the heat loss of the metal ($m= 3.6~\rm kg$). The initial temperature of the water+container is $T_i= 16~\rm C$ and the final temperature $T_f = 18 ~\rm C$. The initial temperature of the small metal was $T_m = 180~\rm C$.

$$(T_f-T_i)(M_c c_m +M_w c_w) = (T_m-T_f)mc_w~.$$

If you solve this equation for $c_m$, you get $c_m = 0.098 ~\rm cal/g\cdot C$.

5-2

The intensity of the sunlight: $I =\rm 700~W/m^2$. Mass of the water: $M=200~\rm kg$. Specific heat of the water: $c = \rm 1~ cal/g\cdot C$. Temperature increase: $\Delta T = 20~\rm C$. Duration of exposure: $\Delta t = 1.0~\rm h = 3600 s$. Efficiency: $\gamma = 0.2$. The unknown area of the collector: $A$.

Then we can set up an equation of the total received energy and the heat gain by the water:

$$I\Delta t \gamma A = \Delta T cM~.$$

So $A={\Delta T cM\over I\Delta t \gamma} = 33.2~\rm m^2$

5-3

Latent heat of the water: $L_v = 2.26\times 10^6 ~\rm J/kg$ (vapourisation) and $L_f = 3.33\times 10^5 ~\rm J/kg$. Specific heat of water: $c = 1.0 ~\rm cal/g\cdot C$. Mass of the ice: $M_i = 150~\rm g$. The unknown mass of the steam: $M_s$. The temperature difference for both: $\Delta T = 50~\rm C$.

The gain and the loss of heat by the ice and the steam are equal:

$$M_i( \Delta T c + L_f) = M_s( \Delta T c + L_v)~.$$

So $M_s = 32.9 ~\rm g$.