What should be the final tempere of the mixture when 10 gm of ice at -10c is mixed with 20 gm of water?
Answers
Answered by
0
The enthalpy of fusion of ice is 334 J/g. The specific heat of water is 4.2 J/g.
To cool 50 g of water from 50 °C to 0 °C would require the removal of
4.2 x 50 x 50 =10500 J.
To melt the ice would require the addition of
334 x 10 = 3340 J
10500 > 3340 thus you can melt all the ice and have some heat to spare, specifically 10500 - 3340 = 7160 J
Now use this to warm up 10 + 50 = 60 g of water at 0 °C
7160 / (60 x 4.2) = 28.4 °C
You will get a slightly different number if you use 4.184 instead of 4.2 J/g.
Similar questions