Question : How much steam at 100°C, will just melt down 3200 g of ice at -10°C?
Given :
Specific heat of ice = 2100 J kg-1°C-1
Specific heat of water = 4200 J kg-1°C-1
Latent Heat of vaporization of steam = 2260×103 J/kg
Latent Heat of fusion of ice = 336×103 J/kg
Doubt by Jaskirat
Solution :
Mass of ice (m) = 3200 g = 3200/1000 = 3.2 kg
Let required mass of steam be m kg
Let required mass of steam be m kg
Amount of Heat gained by 3.2 kg of ice at -10°C to get converted into water at 0°C
=Heat to required to convert 3.2 kg of ice at -10°C to ice at 0°C (Q1) + Heat required to convert 3.2 kg of ice at 0°C to water at 0°C (Q2)
Mass of ice (m) = 3200 g = 3200/1000 = 3.2 g
Q1=mcΔT
Q1=(3.2)(2100)[0-(-10)]
Q1=32×2100
Q1=67200 J
Q2=mL
Q2=3.2(336×103)
=Heat to required to convert 3.2 kg of ice at -10°C to ice at 0°C (Q1) + Heat required to convert 3.2 kg of ice at 0°C to water at 0°C (Q2)
Mass of ice (m) = 3200 g = 3200/1000 = 3.2 g
Q1=mcΔT
Q1=(3.2)(2100)[0-(-10)]
Q1=32×2100
Q1=67200 J
Q2=mL
Q2=3.2(336×103)
Q2=1075200
Amount of Heat lost by m kg of steam at 100°C to convert into water at 0°C = Heat required to convert m kg of steam from 100°C to water at 100°C (Q3) + Heat required to convert m kg of water at 100°C to water at 0°C (Q4)
Q3=mL
= m(2260×103)
=2260000m
Q4=mcΔT
Q4=m(4200)(100-0)
Q4=420000m
As per the principle of calorimetry
Heat Gained = Heat Lost
Q1+Q2=Q3+Q4
Q3=mL
= m(2260×103)
=2260000m
Q4=mcΔT
Q4=m(4200)(100-0)
Q4=420000m
As per the principle of calorimetry
Heat Gained = Heat Lost
Q1+Q2=Q3+Q4
67200+1075200=2260000m+420000m
1142400=2680000m
1142400/2680000 = m
m = 0.4262 kg
m = 426.2 g (approx)
Hence, required amount of steam = 426.2 grams
1142400=2680000m
1142400/2680000 = m
m = 0.4262 kg
m = 426.2 g (approx)
Hence, required amount of steam = 426.2 grams