course Phy 202
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I think I may have finally figured this one out-- How much ice at 0C should we add to 220 grams of water at 35C and 16 grams of steam at 100C to end up with only water at 0C, assuming that all thermal exchanges take place within this system? For the first part 220g of water at 35C to 0C, I used Q=mc'dt=.22kg(4186J/kgC)(-35C)=-32232.2J/gram.
Good but the unit of kg * J / (kg C) * C is just J, not J / gram.
For the second part 16grams of steam to 0C, Q=.016kg(2010J/kgC)(-100C)=-3216J/gram.
Again, it comes out just Joules, not J / gram.
Since the water and steam will lose 35448.2J/gram, the ice must give that much up. 35448.2J*1gram/335J=105.8 grams of ice are required.(35448.2J/gram just J
came from -32232.2+ -3216). Is this correct?
Good, except you didn't include the thermal energy lost when the 16 g of steam condenses (the steam loses 540 cal / gram, at 4.19 J for every cal). That's going to about double your final answer.