Answer:
Total heat required to raise the temperature of water from 45.7°C to 103.5°C
= 249,362.4 J
Explanation:
The Heat required to raise the temperature of 100.0 g of water from 45.7°C to 103.5°C will be a sum of;
- The heat required to raise the 100 g of water from 45.7°C to water's boiling point of 100°C
- The Heat required to vaporize the 100 g of water at its boiling point
- The Heat required to raise the temperature of this vapour from 100°C to 103.5°C
1) The heat required to raise the 100 g of water from 45.7°C to water's boiling point of 100°C
Q = mCΔT
m = 100 g
C = 4.18 J/g.°C
ΔT = change in temperature = (100 - 45.7) = 54.3°C
Q = 100 × 4.18 × 54.3 = 22,697.4 J
2) The Heat required to vaporize the 100 g of water at its boiling point
Q = mL
m = 100 g
L = ΔHvaporization = 2260 J/g
Q = mL = 100 × 2260 = 226,000 J
3) The Heat required to raise the temperature of this vapour from 100°C to 103.5°C
Q = mCΔT
m = 100 g
C = 1.90 J/g.°C
ΔT = change in temperature = (103.5 - 100) = 3.5°C
Q = 100 × 1.9 × 3.5 = 665 J
Total heat required to raise the temperature of water from 45.7°C to 103.5°C
= 22,697.4 + 226,000 + 665
= 249,362.4 J
Hope this Helps!!!
