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).

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Infinity Learn · Accepted Answer

loss in Potential Energy $$=$$ Heat Energy produced $$=$$ Q and Q $$=$$ ms∆t .Loss in Potential Energy $$=mg(H$$−$$h)=ms$$Δ$$t10(10$$−$$5.6)=460$$×Δ$$t10$$×$$4.6=460$$×ΔtΔ$$t=0.1$$∘C

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).

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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).

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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).