2
$\begingroup$

I know how to prove that the efficiency is $1-\frac{Q_c}{Q_h}$, but how do I go from that to $1-\frac{T_c}{T_h}?$ i.e. what is the proof that $\frac{Q_c}{Q_h}=\frac{T_c}{T_h}$?

$\endgroup$

closed as off-topic by Danu, John Duffield, ACuriousMind, Kyle Kanos, Martin Mar 11 '16 at 13:34

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Danu, John Duffield, ACuriousMind, Kyle Kanos, Martin
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 3
    $\begingroup$ You can't without assuming an ideal heat cycle. $\endgroup$ – Viktor Mar 10 '16 at 13:10
3
$\begingroup$

Consider a Carnot heat cycle as given below:

P-V Diagram

Now the Equation of efficiency, denoted by $\boldsymbol \eta$ for a Carnot cycle is given by:

$$\boldsymbol \eta = \frac{\boldsymbol W}{\boldsymbol Q_R} = \frac{\boldsymbol Q_H - \boldsymbol Q_c}{\boldsymbol Q_H} = \boldsymbol 1-\frac{\boldsymbol Q_C}{\boldsymbol Q_H}$$

In an isothermic process,

$\Delta U = 0$;

$Q=W=nRT \log \frac{\boldsymbol V_2}{\boldsymbol V_1}$

For Isothemic process $A \rightarrow B$ and $C \rightarrow D$

$${\boldsymbol Q_c=nRT_c \ln \frac{V_B}{V_A}}$$

$${\boldsymbol Q_H=nRT_H \ln \frac{V_D}{V_C}}= - {nRT_H \ln \frac{V_C}{V_D}}$$

Now by Dividing $\boldsymbol Q_c$ by $ \boldsymbol Q_H$, we have

$$\frac{Q_c}{Q_H}= \frac{nRT_c \ln \frac{V_B}{V_A}}{-nRT_H \ln \frac{V_C}{V_D}} = \frac{T_c \ln \frac{V_B}{V_A}}{-T_H \ln \frac{V_C}{V_D}} \tag{1}$$

Now from Adiabatic processes $B \rightarrow C$ and $D \rightarrow A$

$$T_CV_B^{\gamma-1} = T_HV_C^{\gamma-1} \tag{2.1}$$

$$T_HV_D^{\gamma-1} = T_CV_A^{\gamma-1} \tag{2.2}$$

Now by dividing $eq-(2.1)$ by $eq-(2.2)$, we have

$$\frac {T_CV_B^{\gamma-1}}{T_CV_A^{\gamma-1}} = \frac {T_HV_C^{\gamma-1}}{T_HV_D^{\gamma-1}}$$

$$\frac {V_B^{\gamma-1}}{V_A^{\gamma-1}} = \frac{V_C^{\gamma-1}}{V_D^{\gamma-1}}$$

Which means, $$\frac {V_B}{V_A} = \frac{V_C}{V_D} \tag{3}$$

Now the final part:

putting relation from $eq-(3)$ in $eq-(1)$, we have

$$\frac{Q_c}{Q_H}= - \frac {T_C}{T_H}$$

By taking the absolute values,

$$\color{green} {\left| \frac {Q_c}{Q_H} \right| = \frac {T_C}{T_H}}$$

Now we can say that,

$$\boldsymbol \eta = 1-\frac{Q_C}{Q_H} = 1-\frac{T_C}{T_H}$$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.