QUESTION: Liquid water at 0 degrees celcius and atmospheric pressure has a densi

Question

QUESTION: Liquid water at 0 degrees celcius and atmospheric pressure has a density of 1g/cm^3. At the same conditions, ice has a density of .917g/cm^3. How much work is done at these conditions by 1 kg of ice as it melts to liquid water?

REMARKS: Please explain in detail if there are any tricky things in this problem. Also explain in great detail the formula used. In other words explain what each symbol or letter in the formula stands for. Also if there are any conversions done, please explain them as well. I really want to know what I'm doing.

thanks, eddie

Explanation / Answer

Work done =PV Where P is the atmospheric pressure V= change in volume. Volume of the ice =1kg/917 Kg/m3=1.09 x 10-3 m3 Volume of liquid water = 1kg/1000 Kg/m3=0.001 m3 V=1.09x10-3 -0.001=0.00009 m3 Atmospheric pressure P=101325 N/m2 Work done = 101325 x 0.00009 = 9.12 J Work done = 101325 x 0.00009 = 9.12 J

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