Question

A force of 22.7 N stretches an elastic band at room temperature. The rate at which its entropy changes as it stretches is about _____ J/Km. Round your answer to 3 decimal places.

Answer 1

Answer:

The value is  $\frac{\Delta S }{ L} = - 0.0721 \ J / km$

Explanation:

From the question we are told that

   The  force is  $F = 22.7 \ N$

    The value of room temperature is $T = 298 \ K$

Generally the rate at which its entropy changes as it stretches is mathematically represented as

         $\frac{\Delta S }{ L} = - \frac{F}{T}$

=>      $\frac{\Delta S }{ L} = - \frac{21.5}{ 298 }$

=>      $\frac{\Delta S }{ L} = - 0.0721 \ J / km$