A metal bar having a mass of 212 g is heated to 125.0 C and then dropped into 375 g of water at 24.0 C. After a while the temp of the water and the metal is 34.2 C. If the specific heat of the metal is 0.831 J/g C, how much heat did the metal lose? How much heat did the water absorb?

Question

Here is my work so far:

212 X 0.831 = 176.172 X 34.2

I am missing a step but I don't know what it is. can someone please help me?

Thank you, Evan

Answer 1

Here is my work so far:

212 X 0.831 = 176.172 X 34.2

q = mass metal x specific heat metal x (Tfinal-Tinitial)
q = 212 x 0.831 x (125.0-34.2) = ?

q gained by water =
mass H2O x specific heat H2O x (Tfinal-Tinitial) = ?

By the way, I don't like the way you wrote your equation.
You have

212 X 0.831 = 176.172 X 34.2 and that isn't true. I see what you've done but the math isn't right.
212 x 0.831 is 176.172 BUT THAT ISN'T THE SAME AS 176.172 X 34.2. If you want to do it that way you should do it in two steps; i.e.,
212 x 0.831 = 176.172, then
176.171 x 34.2 = ?
However, you can save a lot of time by not copying that intermediate number down. Do it this way.
212 x 0.831 x 90.8 and do the two multiplications in sequence. Leave that intermediate number in your calculator and you never have to re-enter it.