Question

(b) for this maximum work, how much heat is exhausted to the cold reservoir per cycle?
a) if cyclic
$e(max)=\frac{w}{varphi_h}=1 - \frac{t_c}{t_h}=1-\frac{(273 + 25)}{(273+230)} = 0.7$
$rightarrow w(max)=e(max)varphi_h$
$=?\times1150 j=? j$
$e_m: not change$
b) $varphi_c =?$ if cyclic
$e_m=\frac{w}{varphi_h}=\frac{varphi_h-varphi_c}{varphi_h}rightarrowvarphi_h-varphi_c = evarphi_h$
$varphi_c=(1 - e)varphi_h$
$rightarrowvarphi_h=varphi_c + w$
$maxrightarrowvarphi_c=varphi_h - w=(1150 -?)=? j$
$e=\frac{w}{varphi_h}$
if cyclic
$e(max)=\frac{w}{varphi_h}=1-\frac{t_c}{t_h}$

Explanation:

Step1: Calculate maximum efficiency

The formula for the maximum efficiency $e_{max}$ of a heat - engine is $e_{max}=1-\frac{T_c}{T_h}$, where $T_c = 273 + 25=298\ K$ and $T_h=273 + 230 = 503\ K$. So, $e_{max}=1-\frac{298}{503}\approx0.4076$.

Step2: Calculate maximum work

Given $\Phi_h = 1150\ J$, and $W_{max}=e_{max}\Phi_h$. So, $W_{max}=0.4076\times1150\approx468.74\ J$.

Step3: Calculate heat exhausted to cold reservoir

For a cyclic process, $e=\frac{W}{\Phi_h}=\frac{\Phi_h-\Phi_c}{\Phi_h}$, and $\Phi_c=(1 - e)\Phi_h$. Substituting $e = e_{max}$ and $\Phi_h = 1150\ J$, we get $\Phi_c=(1 - 0.4076)\times1150=0.5924\times1150 = 681.26\ J$.

Answer:

$681.26\ J$