Hi, here is the problem that I'm having trouble with. It is desired to cool iron parts from 600°F to 200°F by dropping them into water that is initially at 55°F. Assuming that all the heat from the iron is transferred to the water and that none of the water evaporates, how many kilograms of water are needed per kilogram of iron? I've used this equation: mass of iron parts * specific heat of iron * change in temp. of iron parts = mass of water * specific heat of water * change in temp. of water I first calculated the change in temperature of the iron parts and converted it to degree Celsius by 200F - 600F = 400F and then 400F*(5/9), which yields -222. I plugged in the values as thus: m*448 J/(kgC)* -222C = m*4186 J/(kgC)* 222C. I've got a sneaking suspicion that it's because of the right side of my equation dealing with water. I think that somehow my scale is wrong, the 222C. But if all the heat released by the iron is transferred to the water with no loss, then the scale would be correct, wouldn't it?