A block of ice at 0°C is added to a 150g aluminum calorimeter cup which holds 210 g of water at 12°C. If all but 2.0 g of ice melt, what was the original mass of the block of ice?

original mass of the block of ice is 38.34 gram

Explanation:

Given data

cup mass = 150 g

ice temperature = 0°C

water mass = 210 g

water temperature = 12°C

ice melt = 2 gram

to find out

solution

we know here

specific heat of aluminum is c = 0.900 joule/gram °C

Specific heat of water C = 4.186 joule/gram °C

so here temperature difference is dt = 12 - 0 = 12°C

so here heat lost by water and cup are given by

heat lost = cup mass * c * dt + water mass * C * dt

heat lost = 150 * 0.900 * 12 + 210 * 4.186 * 12

heat lost = 12168.72 J

so

mass of ice melt here = heat lost / latent heat of fusion

here we know latent heat of fusion = 334.88 joule/gram

so

mass of ice melt = 12168.72 / 334.88

mass of ice melt is 36.337554 gram

so mass of ice is here = mass of ice melt + ice melt

mass of ice = 36.337554 + 2

mass of ice = 38.337554 gram

so original mass of the block of ice is 38.34 gram