Question

A carnot engine having an efficiency of 1/10 as heat engine is used as a refrigerator if the work done on the system is 10j the amount of energy absorbed from the reservoir at low temperature is

Answer 1

$\huge\mathfrak {Answer:-}$

Effieciency of the Carnot Engine:

$η \: = \: \frac{Work \: done}{Energy \: input}$

$\frac{1}{10} = \frac{10J}{Qinput}$

$η \: = 1 - \frac{Qoutput}{Qinput}$

Now,

$Qinput \: = 100J$

And,

$Qoutput = \frac{9}{10} \times 100J$

$Qoutput = 9 × 10$

$Qoutput = 90J$

Therefore,

The amount of energy absorbed from the reservoir at low temperature is $90J.$

Thus,

The answer is $90\: Joules.$

$\huge{Be\:Brainly}$❤️

Answer 2

hey

here is answer

Effieciency of the Carnot Engine:

$η \: = \: \frac{Work \: done}{Energy \: input}$

$\frac{1}{10} = \frac{10J}{Qinput}$

$η \: = 1 - \frac{Qoutput}{Qinput}$

Now, 

$Qinput \: = 100j$

And, 

$\begin{lgathered}Qoutput = \frac{9}{10} \times 100J \\\end{lgathered}$

Qoutput = 9 × 10Qoutput=9×10 

Qoutput = 90J


hope it helps

thanks