Solution :-
Q1) Amount of heat loss by the water will be absorbed by the ice which causes the melting of ice and then melted ice water raises in the temperature from 0 degree to 19.6 degree C.
So we can write it as
-q = +q
-m*c*delta T = (m ice *delta H fus ice)+(m*c*delta T)
-108.31 g* 4.184 J/gC*(19.6C -25.1C) = (5.35g* delta H fus ice)+(5.35g*4.184J/gC*(19.6 C – 0.0C))
2492.43 J = (5.35g* delta H fus ice) + 438.7 J
2492.43 J – 438.7 J = (5.35g* delta H fus ice)
2053.3 J = (5.35g* delta H fus ice)
2053.3 J / 5.35 g = delta H fus
384 J/g = delta H fus
Therefore the enthalpy of fusion of ice is 384 J/g
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