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.

Respuesta :

Answer:

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

Explanation:

From the question we are told that

   The  force is  [tex]F = 22.7 \ N[/tex]

    The value of room temperature is [tex]T = 298 \ K[/tex]

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

         [tex]\frac{\Delta S }{ L} = - \frac{F}{T}[/tex]

=>      [tex]\frac{\Delta S }{ L} = - \frac{21.5}{ 298 }[/tex]

=>      [tex]\frac{\Delta S }{ L} = - 0.0721 \ J / km[/tex]