Chapter 12.2, Problem 25PP

To determine

The work done in boring the hole.

Answer to Problem 25PP

The work done for boring the hole is $\text{1}\text{.8 kJ}$

Explanation of Solution

Given:

A drill bores a small hole in a 0.40-kg block of aluminum which increases its temperature by $5°\text{C}$ .

Formula Used:

According to First Law of thermodynamics, the change in the thermal energy of an object is,

  $\Delta U=Q-W$ ……. (1)

Where,

Q is the heat supplied to the object

W is the work done by the object.

The thermal energy is given by,

  $\Delta U=mC\Delta T$

Here, m represents the mass of the object,C represents specific heat of the object and $\Delta T$ represents change in temperature

Calculation:

Here, work is done on the object which is the aluminum block. The work done by the block is given by,

  $W_b=-W_d$

Also, there is no external source of heat energy. Thus, $Q=0$ .

Now, the thermal energy for the aluminum block of mass $m=0.4-\text{kg}$ , specific heat $C=897\text{ J/kg}\text{.K}$ and with a change in temperature $\Delta T=5°\text{C}$ is obtained as,

  $\begin{array}{c}\Delta U=0.40\times 897\left(5\right)\\ =1.79\times {10}^3\text{ J}\\ \simeq \text{1}\text{.8 kJ}\end{array}$

Thus, based on (1), the thermal energy is given by,

  $\begin{array}{c}\Delta U=-W_b\\ =-\left(-W_d\right)\end{array}$

And, the work done by the drill is,

  $W_d=1.8\text{ kJ}$

Conclusion:

Therefore, the work done by the drill to bore the hole is 1.8 kJ.