10grm ice at 0 degree Celsius + 1grm steam at 100 degree Celsius. Find mass of left substitute. Sir please briefly explain this type of problems.
Asked by amitjena226
17th January 2018,
10:52 AM
Answered by Expert
Answer:
let x gm of ice melt and the final temperature of mixture is 0°C.
Heat loss of steam from 100°C to 0°C water :- ms×(Lv+100) , where ms is mass of steam in kg Lv is latent heat of vapourization.
Heat gain of x gm of ice = x ×10-3 × Lf, wher Lf is laten heat of fusion of ice
x × Lf = ms×(Lv+100)
Lf = 334 kJ/kg, Lv = 2260 kJ/kg,
hence 6.77 gm of ice melts into water
total water content = 6.77gm from melting of ice+1gm of condensed steam = 7.77gm
left out ice = 3.23 gm
Answered by Expert
17th January 2018,
1:50 PM
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