Question

really really need help w/this problem I think i figured it out but i have to be sure i did it right in order to do others similar to it
please help
thanks


Metal Specific Heat (J/g-C)
Al (s) 0.900
Au(s) 0.129
Cu(s) 0.385
Fe(s) 0.444
Hg(l) 0.139
H2O(l) 4.184
C2H5OH(l) 2.46

A piece of metal at a temp of 95.0oC and a mass of 67.0 g is dropped into a beaker of water containing 100 g at a temp. of 30.0oC. If the final temperature reached is 42.7.0oC, what is the identity of the metal?

the sum of the heats gained is zero.
heatgainedmetal+heatgainedwater=0
95*cm*(30-95)=100*4.18(42.7-30)
1.28
would it be Au?


chemistry - DrBob222, Sunday, October 24, 2010 at 10:40pm
I don't think so.
I think the 95 you have listed should be 67 g (at least from the post) and delta T for the metal is (Tfinal-Tintial). The final T from the problem is 42.7 and the initial is 95 isn't it? Check your numbers in the post. That may be a typo.

THIS IS MY REVISED ANSWER

m1*cm(T1-Tm) = m2*cw(Tm-T2)
67*cm*(95-42.7) = 100*4.18*(42.7-30)
cm = .66
What went wrong? because this isnt one of my options?

Answer 1

I get 1.51 which isn't on of the options either.
67X(Tfinal-Tinitial) + 100 x 4.18 x (Tfinial-Tinitial) = 0
67X(42.7-95) + 100*4.184*(42.7-30) = 0
-3504.1X + 5313.7 = 0
X = 5313.7/3504.1 = 1.52
In fact, I don't see it in this table.
http://en.wikipedia.org/wiki/Heat_capacity#Table_of_specific_heat_capacities