Posted by nate on .
a 1.0 kg sample of metal with a specific heat of 0.50KJ/kgCis heated to 100.0C and then placed in a 50.0g sample of water at 20.0C. What is the final temperature of the metal and the water?
I can't figure this out. Is the answer room temperature???

physical science 
bobpursley,
No. The sum of heats gained is zero.
heat gained by water + heat gained my metal=0
mw*cw*(Tf20)+mm*cm*(Tf100)=0
solve for Tf. 
physical science 
nia,
[1.0Kg][1000g][100CTf][0.50KJ][ 1Kg][1000J] = [50.0g][Tf20C][4.184 J]
[ 1Kg] [ KgC][1000g][ 1KJ] [ gC ]
[1000g][100CTf][0.50J] = [50.0g][Tf20C][4.184 J]
[ gC] [ gC ]
[1000g][100CTf][0.50 J/gC] = [Tf20C]
[ 50.0g] [4.184 J/gC]
[2.39][100CTf] = [Tf20C]
239C2.39Tf = Tf20C
239C + 20C = Tf + 2.39Tf
259C = 3.39Tf
259C = Tf
3.39
76C = Tf