- #1

Stephen Bulking

- 54

- 10

- Homework Statement
- I need help with question 2,5,6,7,9,10. I have posted a word file with all the questions below.

- Relevant Equations
- Entropy

1st and 2nd law of Thermodynamics

Heat engine, Carnot engine, C.O.P, efficiency

Q2:

An engine has power of 26.5 kW, needs 9 kg coal during an hour for energy. The heat capacity of coal is 7800 cal/g. Define the engine’s efficiency.

Qh= mcΔT = 9000x7800xΔT ( stuck)

P=26.5kW(is this the power output of the engine in an hour?)

If only I could find ΔT, thenI would be able to calculate Efficiency e=W/Qh; I might get W from the given P, but still the time interval on which the engine operates is not given...

Q5: ( using the given information from Q4)

e(carnot)= 33.3%

Qc = 1000J (I think so)

e=1-Qc/Qh => Qh=1.499J

e= w/Q =>W=Qhxe=0.499; there is an answer very close ti this number but still I can't accept the 0.003 difference.

Q6:

e(carnot)= 1 - Tc/Th=24/25

P =150kW, again I'm not sure on which time interval does the engine has this power but let's just assume it's a second then ==> P(hour)=125/3

e=24/25=P/W ( I think) ==> then W would be 43.4 J ?

Q7:

W=-P(V1-V2) = -0.8(2-9)x10^-3 =5.6x10^-3

Q9 and 10: I think I got this one but I'm not sure about the units of measurements like whether it should be Pa or atm, L or m3 or cm3

W=nRTxln(Vi/Vf)=3000=8,314Txln(Vi/25)

We can find T using PfVf=nRT. This would leave us with one equation and one unknown Vi which could be easily found through a few simple mathematical operations.

Please Help.

An engine has power of 26.5 kW, needs 9 kg coal during an hour for energy. The heat capacity of coal is 7800 cal/g. Define the engine’s efficiency.

Qh= mcΔT = 9000x7800xΔT ( stuck)

P=26.5kW(is this the power output of the engine in an hour?)

If only I could find ΔT, thenI would be able to calculate Efficiency e=W/Qh; I might get W from the given P, but still the time interval on which the engine operates is not given...

Q5: ( using the given information from Q4)

e(carnot)= 33.3%

Qc = 1000J (I think so)

e=1-Qc/Qh => Qh=1.499J

e= w/Q =>W=Qhxe=0.499; there is an answer very close ti this number but still I can't accept the 0.003 difference.

Q6:

e(carnot)= 1 - Tc/Th=24/25

P =150kW, again I'm not sure on which time interval does the engine has this power but let's just assume it's a second then ==> P(hour)=125/3

e=24/25=P/W ( I think) ==> then W would be 43.4 J ?

Q7:

W=-P(V1-V2) = -0.8(2-9)x10^-3 =5.6x10^-3

Q9 and 10: I think I got this one but I'm not sure about the units of measurements like whether it should be Pa or atm, L or m3 or cm3

W=nRTxln(Vi/Vf)=3000=8,314Txln(Vi/25)

We can find T using PfVf=nRT. This would leave us with one equation and one unknown Vi which could be easily found through a few simple mathematical operations.

Please Help.