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course phy242
October 5, Saturday 730pmchapter 20 hw
20.38
An average sleeping person metabolizes at a rate of about 80W by digesting food or burning fat. Typically. 20% of this energy goes into bodily fuctins, such as cell repair, pumping blood,
and other uses of mechanical energy, while the rest goes to heat. Most eople get rid of all this excess heat by transferring it (by conduction and the flow of blood) to the surface of the body, where it is radiated away.
The normal internal temperature of tht body (where the metabolism takes place) is 37 celsius, and the skin is typically 7 celsius cooler. By how much does the persons entropy change per second due to this heat transfer?
Q_T=80 J
Q_used=64 J
T_1=37 C, 310K
T_2=30 C, 303K
Total entropy=delta S
dS=Q/T--->(64J/310K)+(-64J/303K)= -4.8*10^-3 J/K
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dS = dQ / T, not Q / T.
dQ for one second is -64 Joules.
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Entropy decreases
As a budding mechanical engineer, you are called upon to design a Carnot engine that has
2.00 mol of a monotomic ideal gas as its working substance and operates from a high
temperature reservoir at 500 C. The engine is to lift a 15 kg weight 2.00 m per cycle
, using 500 J o heat input. The gas in the engine chamber can have a minimum volume of 5.00 L during the cycle.
a) draw a pV-diagram.
The pV diagram looks like any other Carnot cycle pV. We start at the highest pressure/lowest volume point in the
graph as step one of the process (pta) it follows through the isothermic expansion with the help from Q_H of 500 J
to point b where it then undergoes adiabatic expansion where T decreases in order for gas to expand without absorbing heat.
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The decrease in T is the result of the quick expansion, which is quick enough that negligible heat gets exchanged during the expansion.
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After that, point C goes through an isothermic compression, where the gas gets compressed at a constant temperature. In order to maintain
that constant temperature, the engine must lose heat, Q_c. From point D back to point a, the gas undergoes adiabatic compression, where T must increase
back to TH because the gas is being compressed with no heat exchange.
Qh=W+Ql
T_h=773K W_engine=m*g*h=294J monotomic ideal gas cp=3/2R cv=5/2R
Qc=Qh-W 500J-294J=206J Q_c=206J
T_c=Th*(Qc/Qh) Tc=773*(-206/500) T_c=318K
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You can't do these calculations with Celsius temperatures.
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C). What is thermal efficiency?
e=1-Tc/Th e=1-318/773 e=.589
D). How much energy does the engine waste per cycle?
Q_c=206J
E). What is the maximum pressure the engine must withstand?
pv=nRT P=(2mol*8.315J/mol*k * 773K)/5*10^-3m^3 P=2.57*10^3 Pa
20.45
An experimental power plant at the Natural energy laboratory of hawaii generates electricity from the temperature gradient of the ocean. The surface and deep water temperatures are 27 and 6 C respectively.
a) what is the maximum theoretical efficiency of this plant?
e=1-Th/Tc 1-279/300=.07 e=7%
b). if the power plant is to produce 210 kW of power, at what rate must heat be extracted from the warm water? At what rate must heat be absorbed by the colt water assume the maximum theoretical efficiency.
W=210kw power Qh=210kw/.07 Qh=3*10^6W Qh-W=Qc Qc=2.8*10^6W
c). The cold water that enters the plant leaves it at a temperature of 10 C. What must be the flow rate of the cold water throught the
system? Give your answer in kg/h and in L/h.
Since there this process is neither isothermal nor adiabatic
Q=mcdT (m=flow) c=4190J/kg*k
m=Q_c/(c*dT) =((2.8*10^6W)/((4190J/kg*K)*(21K)))*3600s/hr
m=601431kg/hr m=6*10^5L/hr
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Your approach is good, but you appear to have done at least some calculations which are inappropriately based on ratios of Celsius temperatures.
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20.48.
A person who has skin of surface area 1.85 m^2 and temperature 30.0C is resting in an
insulated room where the ambient air temperature is 20.0C. In this state, a person gets rid of excess heat
by radiation. By how much does the person change the entropy of the air in this room
each second?
Q=mcdT
dS=Q/T
T_1=30C 303K T_2=20C 293K H=A*e*sigma*T^4--->Q
dS=(1.85m^2*1.00*(5.67*10^-8W/m^2*K^4)(1.0s)(303K)^4)/293K-(1.85m^2*1.00*(5.67*10^-8W/m^2*K^4)(1.0s)(293K)^4)/293K
dS=.397J/K
20.51
A monotomic ideal gas is taken around the cycle shows in fig in the direction shown in the figure. The path for the process c to a i s a straight line in the pv diagram. a calculate Q, W and dU for
each process. a to b b to c and c to a. B What are Q w and du for one complete
cycle? C What is the efficiency of the cycle?
cp=5/2R cv=3/2R R=8.315mol/kg*K
process a---->b
constant pressure volume change. W=PdV Q=nCpdT
Q=ncpdT=cp/R*pa*(va-vb)=2.5*(3.00*10^5Pa)*(.300m^3)=2.25*10^5J
W=pdV= (3.00*10^5Pa)*(.300m^3)=2.25*10^5J
dU=Q-W---->dU=1.35*10^5J
process b---->C
constant volume pressure change (W=0) Q=dU
Q=nCvdP*V=3/2(pc-pb)*vB=1.5*(-2.00*10^5Pa)*(.800m^3)=-2.40*10^5J
W=0
dU=Q
process c----->a
dU=0
Q=W
W=.5*(3.00*10^5Pa+1.00*10^5Pa)(.800m^3-.500m^3)=6.00*10^4J
W=-.60*10^5J
dU=1.05*10^5J
b) What are Q,W, and dU for one complete cycle?
Q=W=.30*10^5J dU=0
C) qin=2.25*10^5J+.45*10^5J=2.7*10^5J
e=W/Qh .30*10^5J/2.70*10^5J
e=.111"
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You're doing well, but be sure to check my notes for some possible errors in detail.
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