How much heat will be required to change 5 lbsof ice at 32degree f into water at 82degree f?

Answer 1

heat of fusion is 333.55 (kJ/kg) rephrasing the question to the metric system: How much heat will be required to change 2.27 Kg of ice at 0 degrees into water at 27 degrees? heat of fusion is 333 * 2.27 = 756 kjoules or 756000 joules (melts the ice) specific heat of water (liquid): cp = 4.185 J/gK (this is given in grams, so we have to convert) or cp = 4185 J/kgK multiplying 4185 * 2.27 * 27 = 256,500 joules total = 256,500 + 756000 = 1012500 joules or 241,800 callories or 960 BTU

Answer 2

80 calories of heat are required to melt one gram of ice without altering the temperature. There are 2267.96 grams in five pounds so you would need 181436.8 calories of heat to melt all that ice without raising the temperature.

Answer 3

5(144 + 1x50) BTU

where 144= latent heat of fusion

1= specific heat

Answer 4

1020.8 kilojoules