Question

Numerical Problems

1. If the specific latent heat of ice is 340,000 J/kg and the specific latent heat of steam is 2,300,000 J/kg,
calculate.

the energy needed to change 3 kg of water at 100°C into steam.​

Answer 1

Answer:

91 .8 * 10^5J

Explanation:

m = 3 Kg

L( ice) = 34×10^4J/kg

L( steam)= 23×10^5 J/kg

Heat required to melt the ice at 0°C into water at 0°C

Q1 = mL(ice) = 3×34×10^4 = 10.2×10^5J

Heat required to convert water at 0°C into water at 100°C

Q2= mS(water)∆T

= 3×4182×(100-0)= 12.54×10^5J

Heat required to convert water at 100°C into steam

Q3= mL(steam) = 3×23×10^5 = 69×10^5J

Total heat required to convert 3 Kg water at 100°C into steam

Q = Q1+Q2+Q3

= 10.2×10^5 + 12.54×10^5+ 69×10^5

= 91.8 ×10^5 J

Hope it is correct...