Chapter 17, Problem 101A

To determine

The increase in temperature during the fall.

Answer to Problem 101A

The increase in temperature during the fall is $9.5°\text{C}$ .

Explanation of Solution

Given:

The mass of copper is $m=1.0\text{ g}$ .

The falling height is $h=1.0\times {10}^3\text{ m}$ .

The velocity of copper is $v=70.0\text{m}/\text{s}$ .

Formula used:

The expression for kinetic energy is

  $KE=\frac{1}{2}m{v}^2$

Here, $m$ is the mass and $v$ is the speed.

The expression for the potential energy is

  $PE=mgh$

Here, $m$ is the mass, $g$ is the acceleration due to gravity, and $d$ is the distance.

The expression for the heat transfer is,

  $Q=mC\Delta T$

Here, $m$ is the mass of an object, $C$ is the specific heat, and $d$ is the difference between the final and initial temperatures.

Calculation:

Consider the acceleration due to gravity as $g=9.8\text{m}/{\text{s}}^{\text{2}}$ .

Consider the specific heat of copper as $C=385\text{J}/\text{kg}°\text{C}$ .

The potential energy of the copper is,

  $\begin{array}{l}E=mgh\\ E=\left(1.0\text{ g}\times \frac{1\text{ kg}}{1,000\text{ g}}\right)\times \left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\times \left(1.0\times {10}^4\text{ m}\right)\\ E=9.8\text{ J}\end{array}$

The final energy of the copper is,

  $\begin{array}{l}{E}_{\text{f}}=\frac{1}{2}m{v}^2\\ E_{\text{f}}=\frac{1}{2}\left(1.0\text{ g}\times \frac{1\text{ kg}}{1,000\text{ g}}\right)\times {\left(70.0\text{m}/{\text{s}}^{\text{2}}\right)}^2\\ E_{\text{f}}=2.4\text{ J}\end{array}$

The heat added to the piece is,

  $\begin{array}{l}Q=\frac{1}{2}\left(E-E_{\text{f}}\right)\\ Q=\frac{1}{2}\times \left(9.8\text{ J}-2.4\text{ J}\right)\\ Q=3.7\text{ J}\end{array}$

The increase in temperature during the fall is,

  $\begin{array}{l}Q=mC\Delta T\\ \Delta T=\frac{3.7\text{ J}}{\left(1.0\text{ g}\times \frac{1\text{ kg}}{1,000\text{ g}}\right)\times \left(385\text{J}/\text{kg}°\text{C}\right)}\\ \Delta T=9.5°\text{C}\end{array}$

Conclusion:

Thus, the increase in temperature during the fall is $9.5°\text{C}$ .