Answer:
The temperature of cold reservoir should be 246.818 K for efficiency of 35%
Step-by-step explanation:
In first case we have given efficiency of Carnot engine = 26 % = 0.26
Temperature of cold reservoir
![T_L=281K](https://img.qammunity.org/2020/formulas/physics/high-school/379zi3gno962mgsesa4rv18w879ujdcegy.png)
We know that efficiency of Carnot engine is given by
![\eta =1-(T_L)/(T_H)](https://img.qammunity.org/2020/formulas/physics/college/cuinwzu8bucappm8nug1a6fi1yrhsbgc7j.png)
![0.26 =1-(281)/(T_H)](https://img.qammunity.org/2020/formulas/physics/high-school/nnq89qhg4hr199esc9owdkmhkxh9db09zz.png)
![T_H=379.72K](https://img.qammunity.org/2020/formulas/physics/high-school/84pjcyzkybx2mmybjtfh4opcicvfnj0p9d.png)
For second Carnot engine efficiency is given as 35% = 0.35
And temperature of hot reservoir is same so
![T_H=379.72K](https://img.qammunity.org/2020/formulas/physics/high-school/84pjcyzkybx2mmybjtfh4opcicvfnj0p9d.png)
So
![0.35=1-(T_L)/(379.72)](https://img.qammunity.org/2020/formulas/physics/high-school/sd3g2eldrv9w42q8jmzsxah5ck06012qku.png)
![T_L=246.818K](https://img.qammunity.org/2020/formulas/physics/high-school/uzavqh8r3zgals0jmyz6ch7ujdp8aigd8d.png)
So the temperature of cold reservoir should be 246.818 K for efficiency of 35%