Question

(II) A 64-kg ice-skater moving at 7.5 m/s glides to a stop. Assuming the ice to be at 0°C and that 50% of the heat generated by friction is absorbed by the ice, how much ice melts?

Short answer

The mass of the ice that melts is $2.7\times {10}^{-3}\mathrm{k}\mathrm{g}$.

Step-by-step solution

Understanding of Latent Heat of Fusion

When the phase of a material changes from solid to liquid or from liquid to gas, the material absorbs some amount of heat energy. The amount of heat required to change the phase of a material of unit mass at constant temperature is termed latent heat.

The amount of heat required to change the phase of a substance of mass m is

$Q=mL$.

The amount of heat required to change the phase of 1.0 kg of a substance from solid to liquid is termed latent heat of fusion$\left(L_{\mathrm{F}}\right)$. The latent heat of fusion of ice is $L_{\mathrm{F}}=3.33\times {10}^5\mathrm{J}/\mathrm{k}\mathrm{g}$.

Given data

Mass of the ice-skater is $m=64\mathrm{k}\mathrm{g}$.

Velocity of ice skater is v = 7.5 m/s.

The temperature of ice is $T_1=0^{\circ }\mathrm{C}$.

Determination of kinetic energy of ice-skater

The ice-skater moves on ice with the velocity v, therefore kinetic energy of the ice-skater is

$\begin{array}{c}KE=\frac{1}{2}m{v}^2\\ =\frac{1}{2}\left(64\mathrm{k}\mathrm{g}\right){\left(7.5\mathrm{m}/\mathrm{s}\right)}^2\\ =1800\mathrm{J}\end{array}$.

Determination of the mass of the ice that melts

Let $m^{\prime }$ be the mass of ice that melts.

When the ice-skater glides on ice to stop, then the whole of its kinetic energy gets converted into heat energy due to friction. 50% of this heat energy is then absorbed by the ice and used completely in melting, that is, in changing the ice into water.

So, the heat energy absorbed by the ice is

$Q=m^{\prime }{L}_{\mathrm{F}}$.

Since 50% of the heat generated is absorbed by the ice, you get the following result:

$\begin{array}{c}\frac{50}{100}\left(KE\right)=Q\\ \frac{50}{100}\left(KE\right)=m^{\prime }{L}_{\mathrm{F}}\\ m^{\prime }=\frac{50}{100}\left(KE\right)\times \frac{1}{L_{\mathrm{F}}}\\ =\frac{1}{2}\left(1800\mathrm{J}\right)\times \frac{1}{3.33\times {10}^5\mathrm{J}/\mathrm{k}\mathrm{g}}\\ =270.3\times {10}^{-5}\mathrm{k}\mathrm{g}\\ =2.7\times {10}^{-3}\mathrm{k}\mathrm{g}\end{array}$

Thus, the mass of the ice that melts is $2.7\times {10}^{-3}\mathrm{k}\mathrm{g}$.