Tuesday

October 13, 2015
Posted by **James** on Tuesday, July 24, 2012 at 7:45pm.

- Physics -
**James**, Tuesday, July 24, 2012 at 7:46pmThe answer is supposed to be 27 degrees C; however, I keep getting 25 degrees C. Could someone check my method?

m(w)c(w)T(iw) + m(s)c(s)T(is) = m(w)c(w)T(f) + m(s)c(s)T(f)

where m(w) is the mass of the water, c(w) is the specific heat of water in cal/(g K), m(s) is the mass of steel, c(s) is the specific heat of steel, T(iw)is the initial temperature of the water, T(is) is the initial temperature of the water, and T(f) is the final temperature of the system.

- Physics -
**bobpursley**, Tuesday, July 24, 2012 at 7:59pmI dont like your method, but it might get the right answer.

The sum of the heats gained is zero.

Mw*cw*(Tfw-Tiw)+Ms*Cs(Tfs-Tis)=0

1000*1*(Tf-15)+2000*.107*(Tf-70)=0

Tf(1000+2000*.107)=15000+70*.107*2000

Tf=29980/1214=24.6C

Now, if you used the specific heat of mild steel as .122 cal/gC, which it is, you would get slightly higher temp.

- Physics -
**James**, Tuesday, July 24, 2012 at 8:11pmI tried using a method similar to yours earlier, but I got around 70 to 80 degrees C when using it, so I tried another method. I get the same answer of 24.6C when using my method, so I'm confused as to why you don't like it.

Also, when I use the specific heat of mild steel as .122 cal/gC, I get 25.79 C, which is closer to 27 degrees C.