Question

A carnot heat engine whose low-temperature reservoir is at $70$ has an effiency of $50$ % it is desired to increase the efficiency to $70$ %. Then:

Answer 1

The carnot efficiency is defined as $e=1-\frac{Tc}{Th}$ where Tc is the temperature of the cold reservoir, and Th is the temperature of the hot (both measured in Kelvin).
$7^{0}C=280K$
So, solving for Th, you find
$\frac{Tc}{Th}=1-e=0.5$
$Th=2Tc=560K$
You want to make $e=0.7$, so you need to solve for the Th that can do this, given that Tc is fixed at $280$.
$0.7=1-\frac{Tc}{Th}$

$0.3=\frac{Tc}{Th}$

$Th=3.333\times 280=933.33K$
So, you need to increase the temperature of the hot source by about $373.33K.$