Question

A steel ball of mass 0.1 kg falls freely from a height of 10m and bounces back to a height of 5.4m from ground. If the dissipated energy in this process is absorbed by the ball, the rise in temperature is (specific heat of steel = 460 j / kg-c )

Answer 1

A steel ball of mass 0.1 kg falls freely from a height of 10m and bounces back to a height of 5.4m from ground. If the dissipated energy in this process is absorbed by the ball, the rise in temperature is (specific heat of steel = 460 j / kg-c )

Answer 2

Answer:

0.1 c

Explanation:

mass = 0.1 kg

given ball falls from a height 10 m

H1 = 10

given ball bounces back to a height of 5.4 m

H2 = 5.4

specific heat of steel = s = 460

therefore,

$mg(h1 - h2) = ms t$

where as t = rise in temperature

$0.1 \times 10(10 - 5.4) = 0.1 \times 460 \times t$

$4.6 = 460 \times t$

$t = \frac{4.6}{46}$

$t = 0.1 \: c$

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