Answer to Question #193616 in Molecular Physics | Thermodynamics for Johannes Steven

Question

(a) The pV-diagram which shows a cycle of a heat engine that uses 0.250 mol of an ideal gas with g = 1.40. Process ab is adiabatic. (i) Find the pressure of the gas at point a. (ii) How much heat enters this gas per cycle, and where does it happen? (iii) How much heat leaves this gas in a cycle, and where does it occur? (iv) How much work does this engine do in a cycle? (v) What is the thermal efficiency of the engine?

(b) Freezer has a coefficient of performance of 2.40. The freezer is to convert 1.80 kg of water at 25 degrees Celsius to 1.80 kg of ice at -5.0 degrees Celsius in one hour. (i) What amount of heat must be removed from the water at 25 degrees Celsius to convert it to ice at -5.0 degrees Celsius? (ii) How much electrical energy is consumed by the freezer during this hour? (iii) How much wasted heat is delivered to the room in which the freezer sits?

Answer 1

a) diagram must be

(i) $P_c = 1.5 \ atm , V_c=0.002m^3 , V_a=0.002m^3$

process ac is isochoric

$\therefore \frac{P}{T}=const.$

since ,$P_cV_c=nRT_c$ (1m³=1000 litre)

$T_c=\frac{P_cV_c}{nR}=\frac{1.5\times0.002\times1000}{0.25\times0.0821}=146.16k \\ \boxed{T_c=146.16k}$

process bc is isobaric

$\therefore$ $V\over T$ = cont.

$T_b= \frac{V_bT_c}{V_c}=657.72 k$

process ab is adiabatic

$TV^{\gamma-1}=const.$

so,

$T_a= T_b(\frac{V_b}{V_a})^{\gamma-1}=1200.4k$

Now, since ca is isochoric

$p\over T$ = const.

so , $p_a=\frac{P_cT_a}{T_c}=\frac{1.5\times 1200.4}{146.16}$

$\boxed{P_a=12.32 \ atm}$ pressure of gas at point a.

ii)

process a to b

Q_ab=0 since process is adiabatic

U_ab=nC_v(T_b-T_a)

u_ab=0.25$\times$ $5R\over2$ $\times(657.72-1200.4)$

U_ab=-2819.9 J

Q_ab=U_ab+W_ab

W_ab=-U_ab=2819.9J

process b to c

$U_bc=nC_v(T_c-T_b)=-2658.19J$

$W_{bc}=P(V_c-V_b)=-1050 J$

$Q_{bc}=U_{bc}+W_{bc}$

$Q_{bc}=-3708.19J$

process c to a

$U_{ca}=nC_v(T_a-T_c)=5478.09J$

$W_{ca}=0$ since volume is constant

$\therefore$ $Q_{ca}=5478.09$

$\therefore$ Head enters per cycle , $Q= Q_{ca}=5478.09 J$ (happens in process ca)

iii)

head leaves per cycle , $\boxed{Q= Q_{bc}= +3708.19 J}$ happens in process bc

iV)

$W=W_{ab}+W_{bc}+W{ca}$

$W= 2819.9-1050+0$

$\boxed{W=1769.9 J}$ work done in cycle

v)

$\eta = \frac{W}{Q_{ca}}=\frac{1769.9}{5478.09}= 0.323$

$\boxed{\eta = 0.323 \ or \ 32.3\%}$

b)

given that

C_w=4190 J/Kg.K , Lf=3.34$\times$ 105 J/Kg.K

& Cice = 2010 J/Kg.K

mass of warer M_w=1.80 kg

mass of ice M_ice=1.80 kg

coefficient of performance K = 2.40

$\Delta T_m=-5.0^oC \\ \Delta _{water}=25^oC$

i) the amount of heat removed from the water at 25^oC it to ice at -5^oC

$Q_c =mC_m\Delta T_{ice}-mL_f+mCw\Delta T_{water}$

$Q_c=(1.80kg)[(2010J/Kg.K)(-5.0^oC)-3.34\times 105J/Kg.K+4190J/Kg.K(-25^oC)]$

$\boxed{Q_c=-8.08\times 10^3J}$

the negative energy is sign indicates heat removed from the water

ii) energy consumed by the freezer during this hour gives work done

$W= \frac{Q_c}{K}=\frac{8.08\times 10^3J}{2.40}=3.36\times10^5 J$

ii) Wasted heat delivered from the room is

$|Q|=|Q_c| +W$

= $8.08\times 10^5J + 3.36\times 10^5 J$

=$11.44\times 10^5 J$