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?

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

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.

To find the amount of heat lost by the metal, you can use the equation:

Q = mcΔT

Where:
Q is the amount of heat transferred (in joules),
m is the mass of the metal (in grams),
c is the specific heat capacity of the metal (in J/g°C), and
ΔT is the change in temperature (in °C).

In this case, the mass of the metal (m) is given as 212 g, the specific heat capacity (c) is given as 0.831 J/g°C, and the change in temperature (ΔT) is the difference between the initial temperature (125.0°C) and the final temperature (34.2°C).

Thus, the calculation to find the heat lost by the metal is:

Q = 212 g × 0.831 J/g°C × (34.2°C - 125.0°C)

Please note that the change in temperature should be negative since the metal is losing heat.