First slide

Calorimetry

Question

A block of ice of mass M=13.44 kg slides on a horizontal surface. It starts with a speed v=10.0 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.

Moderate

Solution

Work done by friction,
$W = \frac{1}{2} M v^{2} = \frac{1}{2} \times 13.44 \times (10)^{2} = 672 J = \frac{672}{4.2} cal$
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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