A block of ice of mass M $$=$$ $$13$$ $$.44kg$$ slides on a horizontal surface. It starts with a speed v $$=$$ $$5.0$$ $$m/s$$ and finally stops after moving a distance s $$=$$ $$30$$ m. What is the mass of ice (in$$ $$grams) that has melted due to friction between block and surface? Latent heat of fusion of ice is Lf $$=$$ $$80$$ $$cal/g$$.

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Infinity Learn · Accepted Answer

Work done by friction,       $$W=12Mv2=12$$×$$13.44$$×$$(10)2=672JIf$$ m mass of ice will melt, we have $$6724.2=m(80)The$$ mass of ice that has melted $$m=67280$$×$$4.2=2.0g$$

Calorimetry

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A block of ice of mass M = 13 .44kg slides on a horizontal surface. It starts with a speed v = 5.0 m/s and finally stops after moving a distance s = 30 m. What is the mass of ice (in grams) that has melted due to friction between block and surface? Latent heat of fusion of ice is L_f = 80 cal/g.

Work done by friction,
$W = \frac{1}{2} {Mv}^{2} = \frac{1}{2} \times 13 .44 \times (10)^{2} = 672 J$
If m mass of ice will melt, we have $\frac{672}{4 .2} = m (80)$
The mass of ice that has melted $m = \frac{672}{80 \times 4 .2} = 2 .0 g$

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