how much heat is required to change 400g of ice at -20c into steam at 120c
2 answers
heat= massice*Heatfusion+masswater*c*100+ masswater*Heatvaporization + masssteam*cs*20
Add the heat required for five separate processes:
(1) Heating the ice from -20 to 0 C
(2) Melting the ice at 0 C
(3) Heating the water to 100 C
(4) Evaporating the water at 100 C
(5) Heating the steam from 100 to 120 C.
For (1) and (5) you will need the specific heats of ice and steam. Look them up.
For (2), use the heat of fusion of 80 calories per gram. Multiply by the 400 g. You require 32,000 calories
For (5), use the heat of vaporization of vaporizing the liquid water, 540 cal/g
For (3), the heat required is 400*1*100 = 40,000 calories.
(1) Heating the ice from -20 to 0 C
(2) Melting the ice at 0 C
(3) Heating the water to 100 C
(4) Evaporating the water at 100 C
(5) Heating the steam from 100 to 120 C.
For (1) and (5) you will need the specific heats of ice and steam. Look them up.
For (2), use the heat of fusion of 80 calories per gram. Multiply by the 400 g. You require 32,000 calories
For (5), use the heat of vaporization of vaporizing the liquid water, 540 cal/g
For (3), the heat required is 400*1*100 = 40,000 calories.
1 answer