# Specific Heat Capacities/Latent Heat

1. Apr 9, 2006

### lando45

OK, my teacher set me this question in preparation for my exams, but I was ill when he taught this topic back in November, so I don't really know how to go about answering it. I've tried conducting some research on the web but it hasn't really gotten me any further.

Suppose that water at room temperature of 28°C is put into an ideal refrigerator that maintains an inside temperature of -3°C. The specific heats of water and ice are, respectively, 4187 J/(kg·C°) and 2090 J/(kg·C°); the latent heat of fusion for water is 3.34x105 J/kg.

8. Apr 9, 2006

### lightgrav

looks right ... but you MUST keep better track of what UNITS you have ...

4187 J/kgK . . . 1180878 J/K . . . 334,000 J/kg . . . 19,493,395 J . . .
. . . 3,600,000 J/kWhr . . . 0.1 $/kWhr . . . to avoid this kind of error. 9. Apr 9, 2006 ### lando45 OK I tried$0.54 and that's wrong. Anyone know why?

10. Apr 9, 2006

### lightgrav

(5064595 + 14428800 + 270864) J = 19764259 J /3600000 J/kWhr = 5.49 kWhr
which costs $0.549 . Is your ice still at 0 Celcius? 11. Apr 9, 2006 ### lando45 OK, here are all my calculations so far: 43.2kg x 4187J x 28degrees = 5,064,595.2J 43.2kg x 3.34 x 10^5 = 14,428,800J Total = 14,428,800J + 5,064,595.2J = 19,493,395.2J Convert into kWh: 19,493,395.2 / 3,600,000 = 5.414832 kWh At cost of$0.10 per kWh, total would be: 5.414832 x $0.10 =$0.54

But this is wrong! WHY?!

12. Apr 9, 2006

### lando45

I thought I didn't need to include the ice part? The question asks "What is the cost of making 96 lb (43.2 kg) of ice if electricity costs \$0.10 per kilowatt-hour?" - so surely cooling the water down to 0 degrees then doing the latent thing will make the ice? I don't need the SHC of ice for part a)?

13. Apr 9, 2006