First slide

Specific heat

Question

A steel ball of mass 0.1 kg falls freely from a height of 10 m and bounces to a height of 5.4 m from the ground. If the dissipated energy in this process is absorbed by the ball, the rise in its temperature is (specific heat of steel = 460 JKg^-1K^-1) (g = 10 ms^-2).

Moderate

A

0.01^oC
B

0.1^oC
C

1^oC
D

1.1^oC

Solution

loss in Potential Energy = Heat Energy produced = Q and Q = ms $∆$ t .
Loss in Potential Energy $= mg (H - h) = msΔt$
$\begin{array}{l} 10 (10 - 5 .6) = 460 \times Δt \\ 10 \times 4 .6 = 460 \times Δt \\ Δt = 0 {.1}^{\circ} C \end{array}$

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