Question

How much heat is released when 10 grams of steam at 100 C is cooled and becomes 10 grams of ice at -10 C?

Answer 1

Heat will be released during phase change from steam to water (Q1)during temperature change from 100 degree Celsius to 0 degree celsius(Q2) then phase change from water to Ice (Q3)then again temperature decrease from zero degree celsius to -10 degree Celsius(Q4)

m=10 gram

s= specific heat of water=4.18 J/gram degree Celsius

si=specific heat of ice= 2.10 J/gram degree ceCelsi

Lv= latent heat of vaporization= 2230 J/gram

Lf= latent heat of fusion= 334 J/gram

Q1= mLv = 10*2230 = 22300 J

Q2= ms∆T = 10*4.18*(100-0)= 4180 J

Q3 = mLf= 10*334= 3340 J

Q4= msi∆T= 10*2.10*(10)= 210

Total released heat= Q1+ Q2 + Q3+ Q4

= 30030 J