Question

A 75 -kg block of ice at \(0.0^{\circ} \mathrm{C}\) breaks off from a glacier, slides along the frictionless ice to the ground from a height of $2.43 \mathrm{m},$ and then slides along a horizontal surface consisting of gravel and dirt. Find how much of the mass of the ice is melted by the friction with the rough surface, assuming \(75 \%\) of the internal energy generated is used to heat the ice.

Short answer

(75% of the internal energy generated is used to heat the ice.) Answer: Approximately 0.00405 kg of ice is melted.

Step-by-step solution

Determine the potential energy of ice at 2.43 m height

To find the potential energy of the block of ice, we use the formula: Potential energy (P.E.) = m * g * h Where: m = 75 kg (mass of ice) g = 9.81 m/s² (acceleration due to gravity) h = 2.43 m (height) P.E. = 75 kg * 9.81 m/s² * 2.43 m = 1796.1225 J (Joules)

Calculate the internal energy generated by friction

When the block of ice slides along the rough surface, the potential energy we calculated in step 1 will be transformed into internal energy: Internal energy generated (I.E.G) = P.E. = 1796.1225 J

Determine the internal energy used to heat the ice

Since only 75% of the internal energy generated is used to heat the ice, we have: Internal energy used to heat ice (I.E.H) = 0.75 * I.E.G = 0.75 * 1796.1225 J = 1347.091875 J

Calculate the mass of ice melted

To find the mass of ice melted, we need to use the specific heat and latent heat of fusion for ice. The specific heat of ice (c) is \(2090 \: \mathrm{J/kg} \cdot {\circ} \mathrm{C}\) and latent heat of fusion (L) is \(3.33 \times 10^5 \: \mathrm{J/kg}\). Since the heating happens at \(0.0^{\circ} \mathrm{C}\), we don't have to consider the specific heat of ice for the calculation, only the latent heat of fusion: mass of ice melted (m_melted) = \(\frac{\text{I.E.H}}{\text{L}}\) m_melted = \(\frac{1347.091875 \: \mathrm{J}}{3.33 \times 10^5 \: \mathrm{J/kg}} \approx 0.00405 \: \mathrm{kg}\) So, the mass of ice melted due to friction with the rough surface is approximately 0.00405 kg.