Chapter 14, Problem 62GP

To determine

To find : The rise in temperature of rubber ball.

Answer to Problem 62GP

  $0.141°\text{C}$

Explanation of Solution

Given :

  $V_i=22{\text{ms}}^{\text{-1}}$

  $V_f=12{\text{ms}}^{\text{-1}}$

Formula Used :

  $KE=\frac{1}{2}m{v}^2$

  $Q=ms\Delta T$

Calculation :

Loss of kinetic energy is,

  $KE=\frac{m}{2}\left({22}^2-{12}^2\right)$

  $\Rightarrow \Delta KE=\frac{m}{2}\left(22+12\right)\left(22-12\right)$

  $\Rightarrow \Delta KE=\frac{m}{2}34\times 10=34\times 5\times m$

This is equal to heat lost. So,

  $\Rightarrow m\times 1200\times \Delta T=34\times 5\times m$

  $\Rightarrow \Delta T=\frac{170}{1200}=0.141°\text{C}$

Conclusion :

Hence rise in temperature will be $0.141°\text{C}$