Question

When a ball of mass mis dropped through a height difference ,its potential energy changes by the amount $\mathbf{mg}\mathbf{∆}\mathbf{h}$,where gis the acceleration of gravity, equal to $\mathbf{9}\mathbf{.}\mathbf{81}{\mathbf{ms}}^{\mathbf{-}\mathbf{2}}$. Suppose that when the ball hits the ground, all that energy is converted to heat, increasing the temperature of the ball. If the specific heat capacity of the material in the ball is $\mathbf{0}\mathbf{.}\mathbf{850}{\mathbf{JK}}^{\mathbf{-}\mathbf{1}}{\mathbf{g}}^{\mathbf{-}\mathbf{1}}$, calculate the height from which the ball must be dropped to increase the temperature of the ball by $\mathbf{1}\mathbf{.}{\mathbf{00}}^{\mathbf{0}}\mathit{C}$.

Short answer

The height difference is -86.6 m, and the final height of the ball is 0 m. Hence, the height from which the ball should be dropped is86.6 m.

Step-by-step solution

The given Information.

The acceleration due to gravity is $\mathbf{9}\mathbf{.}\mathbf{81}{\mathbf{ms}}^{\mathbf{-}\mathbf{2}}$.

The specific heat capacity of the material in the ball is ${\text{0.850 JK}}^{\text{- 1}}{\text{g}}^{\text{- 1}}$.

The increase in temperature of the ball must be $1.{00}^{0}C$.

Height from which the ball must be dropped to increase the temperature by 1.000C.

The work done is equated with the heat required to elevate the temperature to obtain the expression for height difference as;

$$\begin{gathered} {\mathbf{\text{- mg∆h = mc}}}_{\mathbf{\text{s}}}\mathbf{\text{∆T}} \\ \mathbf{∆}\mathbf{\text{h = -}}\frac{{\mathbf{\text{c}}}_{\mathbf{\text{s}}}\mathbf{\text{∆T}}}{\mathbf{\text{g}}} \end{gathered}$$

The height from which the ball must be dropped to increase the temperature by .

The total work done is given by the expression;

$\text{W = - mg∆h}$

Here, g denotes the acceleration due to gravity denoted as ${\text{9.81ms}}^{\text{- 2}}$.

The heat required to increase the temperature of the ball is:

${\text{q = mc}}_{\text{s}}\text{∆T}$

The specific heat capacity of the material in the ball is ${\text{0.850 JK}}^{\text{- 1}}{\text{g}}^{\text{- 1}}$.

The change in the temperature of the ball, $∆T=1.{00}^{0}C$.

This is same as 1.00 K as the change is same in any unit.

Since all the energy is converted to heat when the ball hits the ground, so:

$$\begin{gathered} {\text{- mg∆h = mc}}_{\text{s}}\text{∆T} \\ ∆\text{h = -}\frac{{\text{c}}_{\text{s}}\text{∆T}}{\text{g}} \end{gathered}$$

The given values can be substituted in the equation to obtain the height difference:

$$\begin{gathered} \text{∆h = -}\frac{{\text{c}}_{\text{s}}\text{∆T}}{\text{g}} \\ \text{= -}\frac{\left({\text{0.850 JK}}^{\text{- 1}}{\text{g}}^{\text{- 1}}\right)\left(\text{1.00 K}\right)}{\left({\text{9.81 ms}}^{\text{- 2}}\right)}\text{×}\frac{{\text{kgm}}^{\text{2}}{\text{s}}^{\text{- 2}}}{\text{1 J}}\text{×}\frac{{\text{10}}^{\text{3}}\text{g}}{\text{1 kg}} \\ \text{= - 86.6 m} \end{gathered}$$