Chapter 12, Problem 77A

To determine

The amount of the ice block melted as a result of the work done by friction.

Answer to Problem 77A

  $2.0\times {10}^{-5}\text{kg}$

Explanation of Solution

Given:

The mass of block of ice, $m=2.2\text{kg}$

The initial velocity of the ice block, $v_i=2.5\text{m/s}$

The final velocity of the ice block, $v_f=0.5\text{m/s}$

Formula used:

The kinetic energy of an object of mass $m$ travelling at a speed of $v$ is given by,

  $KE=\frac{1}{2}m{v}^2$

And, the work done on an object is given by the change in its kinetic energy,

  $W=K{E}_f-K{E}_i$

Where $K{E}_f$ is the final kinetic energy, and $K{E}_i$ is the initial kinetic energy.

And, the heat of fusion is given by

  $H_f=\frac{Q}{m}$

Where $Q$ is the quantity of heat, m is the mass of the substance.

Calculation:

Heat released in the given process is,

  $\begin{array}{c}Q=-W=-\Delta KE\\ =-\frac{1}{2}m\left(v_f^2-v_i^2\right)\\ =-\frac{1}{2}\left(2.2\right)\left({\left(0.5\right)}^2-{\left(2.5\right)}^2\right)\\ =6.6\text{}\text{J}\end{array}$

Now, it is known that heat of fusion of water is,

  $H_f=3.34\times {10}^5\text{J/kg}$

Then using the mentioned formula, the mass of ice melted is,

  $\begin{array}{c}m=\frac{Q}{H_f}\\ =\frac{6.6}{3.34\times {10}^5}\\ \approx 2.0\times {10}^{-5}\text{kg}\end{array}$

Conclusion:

Thus, the mass of the melted ice is $2.0\times {10}^{-5}\text{kg}$ .