Chapter 18, Problem 110A

To determine

The number of times to drop the bag to heat it by the given temperature.

Answer to Problem 110A

The number of times to drop the bag to heat it by $10°\text{C}$ is $177\text{ times}$ .

Explanation of Solution

Given:

The mass of the bag is $m=2.0\text{ kg}$ .

The height is $h=1.5\text{ m}$ .

The temperature change is $\Delta T=10°\text{C}$ .

Formula used:

The expression for the gravitational 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 lead as $C=130\text{J}/\text{kg}°\text{C}$ .

The gravitational potential energy is,

  $\begin{array}{l}PE=mgh\\ PE=\left(2.0\text{ kg}\right)\times \left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\times \left(1.5\text{ m}\right)\\ PE=29.4\text{ J}\end{array}$

The heat transfer to the bag is,

  $\begin{array}{l}Q=mC\Delta T\\ Q=\left(2.0\text{ kg}\right)\times \left(130\text{J}/\text{kg}°\text{C}\right)\times \left(10°\text{C}\right)\\ Q=2,600\text{ J}\end{array}$

The number of times to drop the bag is,

  $\begin{array}{l}N=\frac{Q}{\frac{1}{2}PE}\\ N=\frac{2,600\text{ J}}{\frac{1}{2}\times 29.4\text{ J}}\\ N=177\text{ times}\end{array}$

Conclusion:

Thus, the number of times to drop the bag to heat it by $10°\text{C}$ is $177\text{ times}$ .