Question

A 2.7 kg block of ice at a temperature of 0.0◦C and an initial speed of 6.5 m/s slides across a level floor. If 3.3 × 105 J are required to melt 1.0 kg of ice, how much ice melts, assuming that the initial kinetic energy of the ice block is entirely converted to the ice’s internal energy? Answer in units of kg.

Answer 1

Answer:

The mass of ice melted is: $1.728*10^{-4}(Kg)$

Explanation:

We need to remember that the latent heat of fusion is related as:$L_{f} =\frac{Q}{m}$, where Lf is the latent heat of fusion, Q is the amount of heat and m is the mass, and when there is a change of phase, e.g ice into liquid water, the temperature remains constant during this process. Now we calculate the kinetic energy of the block as: $E_{k}=\frac{m*v^{2} }{2}=\frac{2.7*6.5^{2} }{2} =57.02(Joules)$, so this energy is used to change ice into liquid water, and knowing $L_{f}=3.3*10^{5}(Joules/Kg)$, then we can replace in the equation the latent heat of fusion and get the mass of ice melted as:$m=\frac{L_{f} }{Q} =\frac{57.04}{3.3*10^{5} } =1.728*10^{-4}(Kg)$