The maximum possible efficiency of a heat engine which exhaust its heat at a temperature of 46.0o is 39.0 percent. What is the minimum value of the temperature at which the engine takes in heat?
*I think I keep doing a step wrong in here so step by step instruction would be so helpful!
Thanks in advance :)
If the engine takes in an amount of heat Q at temperature Tin, the entropy of the environment will decrease by:
-Q/Tin
If the engine performs an amount of work W, then in needs to dump an amount of heat of Qout = Q -W to make the energy balance neutral. If it does this at a temperature of Tout, that will contribute to an entropy rise of the environment of:
Qout/Tout = (Q-W)/Tout
The total entropy change of the environment is thus:
Delta S = -Q/Tin + (Q-W)/Tout
If the engine's internal state is the same after each cycle, then Delta S is the total entropy change and this has to be equal or larger than zero. So, we have:
-Q/Tin + (Q-W)/Tout >= 0
W/Q is the efficiency eta, dividing by Q gives:
eta <= 1 - Tout/Tin ------->
Tout/Tin <= 1 - eta ------->
Tin >= Tout/(1 - eta)
To find the minimum value of the temperature at which the engine takes in heat, we can use the Carnot efficiency formula. The Carnot efficiency is given by the equation:
η = 1 - (Tc/Th)
Where:
- η is the Carnot efficiency
- Tc is the absolute temperature of the cold reservoir (where heat is exhausted)
- Th is the absolute temperature of the hot reservoir (where heat is taken in)
In this case, the Carnot efficiency is given as 39.0 percent, which is equivalent to 0.39 in decimal form. The temperature Tc is given as 46.0o.
Let's now solve for Th step by step:
Step 1: Convert the Carnot efficiency to decimal form
0.39 = 1 - (46.0 / Th)
Step 2: Rearrange the equation to solve for Th
46.0 / Th = 1 - 0.39
Step 3: Subtract 0.39 from 1
1 - 0.39 = 0.61
Step 4: Divide 46.0 by the result from Step 3
46.0 / 0.61 = 75.41
Therefore, the minimum value of the temperature Th at which the engine takes in heat is approximately 75.41o.