An ice cube of mass 100 g and at 0$\degree$C is dropped into a Styrofoam cup containing 200 g of water at 25$\degree$C. (The latent heat of fusion of water is 80 cal/g and the specific heat capacity of water is 1.0 cal/g C$\degree$.) Assuming the cup does not exchange any heat with the outside, what will happen?
A) All of the water will freeze.
B) Some of the water will freeze.
C) Nothing will happen.
D) Some of the ice will melt.
E) All of the ice will melt.

Question

Quizplus · Accepted Answer

To determine the outcome, we calculate the heat exchange between the ice and water. The ice will absorb heat to melt, and the water will cool down. Since the specific heat of water is 1.0 cal/g°C and the latent heat of fusion of water is 80 cal/g, we can calculate the energy needed for the ice to melt and compare it to the energy the water can lose before reaching 0°C. The ice needs 100g * 80 cal/g = 8000 cal to melt completely. The water can lose 200g * 25°C * 1.0 cal/g°C = 5000 cal before reaching 0°C. Since the water doesn't have enough energy to melt all the ice, only some of the ice will melt.

An Ice Cube of Mass 100 G and at 0 °\degree°

An Ice Cube of Mass 100 G and at 0 $\degree$