In: Chemistry
Calculate the amount of heat that must be absorbed by 50.0 grams of ice at -12.0oC to convert it to water at 20.0oC. (specific heat capacity of ice = 2.09 J/g.K; specific heat capacity of liquid water = 4.18 J/g.K; specific heat capacity of water vapor = 1.84 J/g.K; heat of fusion of ice = 334 J/g
This is calculated in 3 stages.
1) To heat the ice from
-12°C to 0°C (dT = 12°C =12K)
Q = mcdT
Q = heat ,
m = mass of water = 50 g
c = specific heat capacity of ice = 2.09 J/g/°C
dT = temperature difference = 12°C = 12 K
Q = 50g x 2.09 J/g/K x 12 K = 1254 J
Q = 1254 J
2) To melt the ice at 0°C
to water at 0°C.
Q = m x heat of fusion of ice
= 50 g x 334 J/g
= 16700 J
Q = 16700 J
3) To heat the water from 0°C to 20°C
Q = mcdT where c = specific heat of capacity liquid water , dT = 20°C = 20 K
= 50 g x 4.18 J/g/K x 20°C
= 4180 J
Q = 4180 J
Total heat to absobed = 1254 J + 16700 J + 4180 J
= 22134 J
Therefore, 22134 J heat must be absorbed by 50.0 grams of ice at -12.0oC to convert it to water at 20.0oC.