# how is it proved $\frac{Q_c}{Q_h}=\frac{T_c}{T_h}$? [closed]

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 $\frac{Q_c}{Q_h}=\frac{T_c}{T_h}$?

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

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

## 1 Answer

Consider a Carnot heat cycle as given below: 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,

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