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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
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,  
begin mathsize 12px style x space equals space fraction numerator m subscript s open parentheses L subscript v plus 100 close parentheses over denominator L subscript f end fraction space equals space fraction numerator 10 to the power of negative 3 end exponent cross times open parentheses 2260 cross times 10 cubed space plus space 100 close parentheses over denominator 334 cross times 10 cubed end fraction space equals space 6.77 space cross times 10 to the power of negative 3 end exponent space k g end style
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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