Question

 answer

A Carnot refrigeration cycle is used to estimate the energy requirement in an attempt to reduce the temperature of a specimen to absolute zero. Suppose that we wish to remove 0.01 J of energy from the specimen when it is at 2x10-6 K. How much work is necessary if the high-temperature reservoir is at 20 0c? ​

Answer 1

Answer:

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Answer 2

Given:

$T_1 = 2\times 10^-^6 K$

$T_2 = 20 deg C$

$Q_1( heat removed) = 0.01J$

To find:

Work done

Solution:

From given data:

$T_2 = 20 deg C = 20+273$

                   $=293K$

According to Carnot's theroem:

$\frac{Q_1}{Q_2}= \frac{T_1}{T_2}$

where $Q_2$ is the energy discharged to body at higher temperature.

Substituting the given values, we get:

$\frac{0.01}{Q_2}= \frac{2\times 10^-^6}{293}$

$Q_2 = 1.465\times 10^6 J$

Now work done, $W=Q_2-Q_1$

                                $= 1.465\times 10^6 - 0.01$

                                $W = 1.464999\times 10^6 J$

Work done will therefore be $W = 1.464999\times 10^6 J$