Question

The specific heat of rubber is 2000 j/kgk. suppose that a rubber ball dropped from a height of 7.0 m bounces back to a height of 3.5 m.

what is the temperature increase of the ball? assume that no energy is transferred to the air or the ground.

Answer 1

Approximately
`0.017\; {\rm K}`, assuming that
`g = 9.81\; {\rm m\cdot s^(-2)}`.

Step-by-step explanation:

This question is asking for the temperature change of the ball associated with a change in gravitational potential energy. While both the change in height
`\Delta h` and the specific heat
`c` of the material are given, the mass
`m` of the ball is not.

The change in the gravitational potential energy of this ball is:

`(\text{change in GPE}) = m\, g\, \Delta h`.

Under the assumptions, all that energy would have been converted into heat. In other words, the
`(\text{GPE})` that has been lost would be equal to the heat energy
`Q` that the ball has absorbed:

`Q = -(\text{change in GPE}) = m\, g\, (-\Delta h)`.

Note that there is a negative sign in front of
`(\text{change in GPE})` in this expression because energy loss is denoted as a negative change, while energy gain is a positive change.

Given the specific heat
`c` of the material, absorbing
`Q = m\, g\, (-\Delta h)` of heat energy would raise the temperature of the material by:

`\begin{aligned}\Delta T &= (Q)/(m\, c) = (m\, g\, (-\Delta h))/(m\, c) = (g\, (-\Delta h))/(c)\end{aligned}`.

Note that the simplified expression for the temperature of the ball does not depend on the mass of the ball.

Substitute in
`g = 9.81\; {\rm m\cdot s^(-2)}`,
`\Delta h = 3.5\; {\rm m} - 7.0\; {\rm m} = (-3.5)\; {\rm m}` (subtract initial value from the new value), and
`c = 2000\; {\rm J\cdot kg^(-1)\cdot K^(-1)}`:

`\begin{aligned}\Delta T &= (g\, (-\Delta h))/(c) \\ &= \frac{(9.81\; {\rm m\cdot s^(-2)})\, (-(-3.5\; {\rm m}))}{(2000\; {\rm J\cdot kg^(-1)\cdot K^(-1)})} \\ &\approx 0.017\; {\rm K}\end{aligned}`.

(Note that
`1\; {\rm J} = 1\; {\rm kg\cdot m^(2)\cdot s^(-2)}`.)

In other words, the temperature of the ball would increase by approximately
`0.017\; {\rm K}`.