# physics

7150/(33.5e4) = g ice doesn't work.
Am I calculating incorrectly?

heat of fusion for water is 334 J/g. You want it in J/g since q is given in the problem in J. The answer will be in units of grams of ice. You will need to change to kg since the problem asks for it in kg.

Should I get 0.0214 kilograms? Because this answer is listed as incorrect.

q=heat of fusion x g ice
You now q. You know heat of fusion (or can look it up). Solve for g ice.

A Carnot engine uses a hot reservoir consisting of a large amount of boiling water and a cold reservoir consisting of a large tub of ice and water. When 7150 J of heat is put into the engine and the engine produces work, how many kilograms of ice in the tub are melted due to the heat delivered to the cold reservoir?

I get the same answer you have. Bob Pursley is on-line but away from his computer at the moment. I will email this to him and let him take a look at it. I don't know exactly when he will be back.

First determine the efficiency:

efficiency= (373-273)/373= .268

That means that 1-.268 is waste heat

Now if 7150J is put into the engine, some is delivered as work, and the rest is waste delivered to the ice.

MassIce melted= (1-.268)7150J/(334J/Kg)

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