A 156.5 g piece of stainless steel was heated to 510 C and quickly added to 25.11 g of ice at -30 C in a well-insulated flask that was immediately sealed. What will be the final temperature of the system if there is no loss of energy to the surroundings? How much liquid will be present when the system reaches its final temperature?
I already have some work but i still can't figure it out..
[massice*specificheatice*(0-(-30))] + [moleice* heatfusionice] + [massH2O*specificheatH2O*(Tf-0)] = -[mass stainless steel*specific heat steel*(Tf-510)]
3 answers
You just solve that equation you have for Tf. There is only the one unknown in the whole thing.
Hold on a minute. I think I gave you a slight bum steer. Let me repost a correction.
No, what I posted, which I presume you obtained from my response to a slightly different problem a little earlier, is ok. The volume of the resultant material (volume of water) was not on the previous post but, assuming a density of 1.00 g/mL for water, the volume will be 25.11 which is the same as the volume of the ice melted. I don't know if the person who presented you with the problem wants you to take into account the difference in densities of water at -30 degrees C and the final T or not but it would be relatively simple to do so.