course Phy 202

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The next problem is How much ice at 0C should we add to 220 grams of water at 35 C and 16 grams of steam at 100 C to end up with only water at 0C, assuming that all thermal exchanges take place within the system? My solution -- for the first part 220g of water at 35C to 0c I used Q=mc'dt = .22kg(4186J/kgC)(-35C)=-32232.2J, for the second part 16 grams of steam at 100 C to 0C -- Q=.016kg(2010J/kgC)(-100C)=-3216J. I am not sure where to go from here or if the above is co Everything you have is a correct contribution to the thermal energy loss of the water and the steam. You did leave off the energy required to condense the steam, which is considerable (on the order of a couple million Joules per kg, but of course you've only got about 1/60 kg of the stuff here). Look up the heat of vaporization (it's 540 calories / gram and a calorie is 4.19 Joules; also easy to remember that the heat of fusion for water is 80 cal / gram). Once you've got that you know how much energy the ice needs to gain. The ice is at 0 C, and ends up as water at 0 C. The energy it gains will be just enough to supply the necessary heat of fusion."