# Calorimetry

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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 ofice (in grams) that has melted due to friction between block and surface? Latent heat of fusion of ice is Lf=80 cal/g.

Moderate
Solution

## Work done by friction,$W=\frac{1}{2}M{v}^{2}=\frac{1}{2}×13.44×\left(10{\right)}^{2}=672\mathrm{J}=\frac{672}{4.2}\mathrm{cal}$If m mass of ice will melt, we have $\frac{672}{4.2}=m\left(80\right)$The mass of ice that has melted $m=\frac{672}{80×4.2}=2.0\mathrm{g}$

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The melting temperature of a certain mass of a solid object is 20°C. When a heat Q is added to the object, its $\frac{3}{4}$material melts. When an additional amount of Q heat is added the material transforms to its liquid state at 50°C. Find the ratio of specific latent heat of fusion (in J/g) to the specific heat capacity of the liquid (in J g-1 °C-1) for the material.