The minimum amount of 100-degree-C steam needed to melt 1 gram of 0-degree-Celsius ice is
A) 0.125 gram.
B) 6.75 grams.
C) 0.148 gram.
D) 8 grams.
E) none of these

Question

Quizplus · Accepted Answer

The heat of fusion of ice is 334 J/g, and the specific heat of water is 4.18 J/g-°C. So, the heat required to melt 1 gram of ice is 334 J, and the heat required to raise the temperature of 1 gram of water from 0°C to 100°C is 4.18 J/g-°C * 100°C = 418 J. So, the total heat required is 334 J + 418 J = 752 J. The heat of condensation of steam is 2260 J/g, so the minimum amount of 100-degree-C steam needed to melt 1 gram of 0-degree-Celsius ice is 752 J / 2260 J/g = 0.125 gram.

The Minimum Amount of 100-Degree-C Steam Needed to Melt 1