Question

A 50 gram holds 400 grams liquid water and 75 grams of ice. After 20 minutes the mass of ice is 50 grams. IN 1-2 sentence, assuming no evaporation or sublimation has occurred, calculate the mass of the liquid water at this time and explain how this is found.

Answer 1

Initially, there were 400 grams of liquid water and 75 grams of ice. After 20 minutes, with 50 grams of ice remaining, the mass of the liquid water must have increased by the mass of ice that melted (since no mass was lost to evaporation or sublimation). Therefore, initially, there were 75 grams of ice minus 50 grams of residual ice, which means 25 grams of ice melted, adding to the liquid water. Thus, the new mass of liquid water is \( 400 \, \text{g} + 25 \, \text{g} = 425 \, \text{g} \).