Question

1. Calculate the difference, to the nearest kJ, in the amount of heat energy required to melt 500g of ice at 0 degrees Celsius and to evaporate 500g of water at 100 degrees Celsius.

2. Calculate the mass of water that could be evaporated at 100 degrees Celsius if 1000 kJ of heat were absorbed by the water.

Answer 1

The heat of fusion for water at 0 °C is approximately 334 joules (79.7 calories) per gram

The heat of vaporization for water at 0 °C is approximately 2260 joules (540 calories) per gram

The amount of heat energy required to melt 500g of ice at 0 degrees Celsius is calculated as follows:

334 joules / gram * 500 grams

= 167000 J

The amount of heat energy required to evaporate 500g of water at 100 degrees is calculated as follows:

2260 joules / gram * 500 grams

= 1130000 J

The difference, to the nearest kJ, in the amount of heat energy required to melt 500g of ice at 0 degrees Celsius and to evaporate 500g of water at 100 degrees Celsius is calculate as follows:

1130000 J - 167000 J

= 963000 J*1 KJ/1000 J

= 963 KJ

Now calculate the mass of water that could be evaporated at 100 degrees Celsius if 1000 kJ of heat were absorbed by the water as follows:

The heat of vaporization for water at 100 °C is approximately 2260 joules (540 calories) per gram

the mass of water = Total heat / heat of vaporization per gram

= ( 1000 kJ *1000 J/1 kj) 2260 joule / g

= 442.5 g