Chapter 12.2, Problem 26PP

To determine

To Calculate: The number of times a bag of lead shot needs to be dropped to heat it by the given temperature.

Answer to Problem 26PP

11

Explanation of Solution

Given:

Mass of bag of lead shot, $m=0.50\text{kg}$

Height from which bag is dropped, $h=1.\text{5}\text{m}$

Required rise in temperature, $\Delta T=1^{\text{o}}\text{C}$

Formula Used:

Heat required to raise the temperature can be obtained by:

  $Q=mc\Delta T$

Where, m is the mass, c is the specific heat capacity and $\Delta T$ is the difference in temperature.

Also, work done when the bag is dropped = gravitational potential energy.

  $W=mgh$

Where, m is the mass, g is the gravitational acceleration and h is the height.

Calculations:

Specific heat capacity of lead, $c=160\text{J/kg}\cdot \text{°C}$

Amount of heat required to raise the temperature of the bag of lead shots by $\Delta T=1^{\text{o}}\text{C}$ is:

  $\begin{array}{c}Q=mc\Delta T\\ =0.50\text{kg}\times 160\text{J/kg}\cdot \text{°C}\times 1\text{°C}\\ \text{=80}\text{J}\end{array}$

Amount of heat generated when bag is dropped once:

  $\begin{array}{l}W=mgh\\ =\left(\text{0}\text{.50}\text{kg}\right)\left(\text{9}\text{.8}{\text{m/s}}^{\text{2}}\right)\left(\text{1}\text{.5}\text{m}\right)\\ =7.35\text{}\text{J}\end{array}$

Thus, the number of times that the bag must be dropped is:

  $\begin{array}{c}n=\frac{Q}{W}\\ =\frac{\text{80}\text{J}}{\text{7}\text{.35}\text{J}}\\ \approx 11\end{array}$

Conclusion:The number of times a bag of lead shot needs to be dropped to heat it by the given temperature is 11.