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HW 13 question 6

0 votes
215 views

67-g ice cube at 0°C is placed in 889 g of water at 26°C. What is the final temperature of the mixture?

I used:

Ewater lost=Eice absorbed+Elatent heat

Mwater*Cwater*(26-T)=Mice*Cice*(T-0)+Mice*Lfusion

And I could not find the correct answer

I used Cwater=4.19*10^3 and Cice=2090

asked Nov 16, 2013 in General by aacc_1234 (240 points)

1 Answer

0 votes
On the right hand side of the equation provided, it means to heat ice to a temperature larger than 0ºC, which is impossible. Ice melts at 0ºC, above which we should be heating water (melted ice part). Therefore, if we replace the specific heat of ice with water, the whole thing should work out.
answered Nov 16, 2013 by PhysicsDude (21,750 points)
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