Question 962: Thermodynamics & KTG

Question

If $\alpha$ is the coefficient of performance of a refrigerator and ' $Q$ ' is heat released to the hot reservoir, then the heat extracted from the cold reservoir ' $Q_2$ ' is

Accepted Answer

To determine the heat extracted from the cold reservoir, $ Q_2 $, in a refrigerator, we use the relationship involving the coefficient of performance $ \alpha $. The coefficient of performance of a refrigerator is defined as:
$  \alpha = \frac{Q_2}{W}  $
where $ W $ is the work done by the refrigerator, given by:
$ W = Q_1 - Q_2 $
Substituting into the formula for $ \alpha $:
$ \alpha = \frac{Q_2}{Q_1 - Q_2} $
Rearranging the equation to express $ Q_2 $ in terms of $ Q_1 $ and $ \alpha $:
$ \alpha (Q_1 - Q_2) = Q_2 $
Simplifying further:
$ \alpha Q_1 - \alpha Q_2 = Q_2 $
Adding $\alpha Q_2$ to both sides and factoring out $ Q_2 $ gives:
$ \alpha Q_1 = Q_2 + \alpha Q_2 $
$ \alpha Q_1 = Q_2 (1 + \alpha) $
Finally, solving for $ Q_2 $:
$ Q_2 = \frac{\alpha Q_1}{1+\alpha} $
This formula shows the relationship between the coefficient of performance and the heat exchanged in the refrigerator system.

Answer

$\frac{\alpha Q_1}{\alpha-1}$

Answer

$\frac{\alpha-1}{\alpha} Q_1$

Answer

$\frac{\alpha Q_1}{1+\alpha}$

Answer

$\frac{1+\alpha}{\alpha} Q_1$

Thermodynamics & KTG

If \alpha is the coefficient of performance of a refrigerator and ' Q ' is heat released to the hot reservoir, then the heat extracted from the cold reservoir ' Q_2 ' is

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