Asked by Anonymous
Two 36.0-g ice cubes initially at 0°C are added to 465 g of water at 22.0°C. Assuming this system is insulated and ignoring heat transfer with the glass, what is the equilibrium temperature of the mixture?
I keep getting 14.7 but it's saying I'm wrong by more than 10%.
I keep getting 14.7 but it's saying I'm wrong by more than 10%.
Answers
Answered by
Anonymous
72 g ice
heat to melt ice = heat of fusion of water * 72 = 80 calories/g *72g = 5760 cal
how much does that cool the water ?
5760 cal = 1 cal /(g deg) * (22-T) 465 g
so 22-T =12.4
T = 9.6 degrees C
NOW we have a new problem
72g water at 0 degreesfrom melted ice and 465 g water at 9.6 degrees C
72(T-0) = 465 (9.6-T)
72 T = 4464 - 465T
T = 4464 / 393 = 11.4 degrees final
heat to melt ice = heat of fusion of water * 72 = 80 calories/g *72g = 5760 cal
how much does that cool the water ?
5760 cal = 1 cal /(g deg) * (22-T) 465 g
so 22-T =12.4
T = 9.6 degrees C
NOW we have a new problem
72g water at 0 degreesfrom melted ice and 465 g water at 9.6 degrees C
72(T-0) = 465 (9.6-T)
72 T = 4464 - 465T
T = 4464 / 393 = 11.4 degrees final
Answered by
Anonymous
Whoops, sign error
72 T = 4464 - 465T
T = 4464 /537= 8.31 degrees final
72 T = 4464 - 465T
T = 4464 /537= 8.31 degrees final
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