Heat of fusion and heat of vaporization
Sample problem:
How many calories of heat are required by 100grams of ice at -50C to change to steam at 1200C?
- Heat to raise temperature of ice to 00C, the melting point (Q1)?
Given:
m = 100g t2 = 0
Sice = 0.5 cal/g 0C t1 = - 50C
Find Q1
Formula Q1 = m s Δt
Substitution: Q1 = (100g) (0.5 cal/g 0C) [0 – (- 50C)]
Answer: Q1 = 250 cal
- Heat to melt ice at O0C (Q2)?
Given:
m = 100g hf = 80 cal/g
Find Q2
Formula Q2 = m hf
Substitution: Q2 = (100g) (80 cal/g)
Answer: Q2 = 8000 cal
- Heat to raise the temperature of water at 00C to 1000C (Q3)?
Given:
m = 100g t2 = 1000C
Swater = 1 cal/g 0C t1 = 0
Find Q3
Formula Q3 = m s Δt
Substitution: Q3 = (100g) (1 cal/g 0C) (1000C-0)
Answer: Q3 = 10,000 cal
- Heat to convert water at 1000C (Q4)?
Given:
m = 100g hv = 540 cal/g
Find Q4
Formula Q4 = m hv
Substitution: Q4 = (100g) (540 cal/g )
Answer: Q4 = 54,000 cal
- Heat to raise the temperature of steam at 1000C to 1200C (Q5)?
Given:
m = 100g t2 = 1200C
Ssteam = 0.48 cal/g 0C t1 = 1000C
Find Q5
Formula Q5 = m s Δt
Substitution: Q5 = (100g) (0.48 cal/g 0C) (1200C- 1000C)
Answer: Q5 = 960 cal
QT = Q1 + Q2 + Q3 + Q4 + Q5
QT = 250 cal + 8,000 cal + 10,000 cal + 54,000 cal + 960 cal
QT = 73,210 cal
Practice problem:
1. How many calories of heat are required by 300grams of ice at -150C to change to steam at 1200C?
2. How many calories of heat are required by 250grams of ice at -330C to change to steam at 1200C?
No comments:
Post a Comment