A 10 kg mass falls through 25 m on to ground and bounces to a height of 0.50 m. Assume that all potential energy lost is used in heating up the mass. The temperature rise will be (specific heat of the material is$\text{252}{\text{Jkg}}^{-1}{\text{K}}^{-1}$ )

Given that

mass of falling body = m

height from which it is falling ${\text{h}}_{\text{1}}=25\text{m}$                         

height to which it is raised after bouncing ${\text{h}}_{\text{2}}=0.50\text{m}$

change in P.E $\text{=}\left({\text{mgh}}_{\text{1}}-{\text{mgh}}_{\text{2}}\right)$

$=\text{mg}\left({\text{h}}_{\text{1}}-{\text{h}}_{\text{2}}\right)$

loss of potential energy = Q produced

m = 10 kg

$\text{S}=252{\text{Jkg}}^{-1}{\text{k}}^{-1}$

$\Delta \text{t}=?$

$10\times 10\times 24.5=10\times 252\times \Delta \text{t}\to \left(1\right)$                             

$\Delta \text{t}=\frac{10\times 10\times 24.5}{10\times 252}=\frac{245}{252}$

$\Delta \text{t}=0.972\text{K}$