Chapter 15, Problem 39P

To determine

To find: The change in entropy.

Answer to Problem 39P

  $0.15{\text{JK}}^{-1}$

Explanation of Solution

Given Data:

  $m=10\text{ kg}$

  $V=3{\text{ ms}}^{-1}$

  $T=293$

Formula Used:

  $KE=\frac{1}{2}m{v}^2\Delta S=\frac{\Delta Q}{T}$

Calculation: change in K.E. = change in heat

  $\begin{array}{l}\Rightarrow \Delta KE=\frac{1}{2}m{v}^2=\frac{1}{2}\times 10\times 9\\ \Rightarrow \Delta KE=45\end{array}$

Change in entropy

  $\begin{array}{l}\Rightarrow \Delta S=\frac{\Delta Q}{T}=\frac{45}{293}\\ \Rightarrow \Delta S=0.15J{K}^{-1}\end{array}$