264.6 g of water is in a Styrofoam calorimeter of negligible heat capacity. The initial T of the water is 16.9oC. After 270.9 g of an unknown compound at 71.3oC is added, the equilibrium T is 33.2oC. What is the specific heat of the unknown compound in J/(goC)?

I did:

270.9(33.2-71.3)+4.184*264.6*(33.2-16.9)=0

-10321.29+18045.50832=0

-18045.51/-10321.29=1.7, but that wasn't right. So I thought maybe I had switched the division so I did -10321.29/-18045.51=.57, but that wasn't right either.

Thanks in advance!

You have it a little mixed up. You don't have specific heat metal at all; you have specific heat H2O added instead of multiplied.

heat lost by metal + heat gained by H2O = 0

[mass metal x specific heat metal x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)

I'm afraid I don't understand what you mean. I had multiplied by the specific heat of water.

Yes, I saw the water.

I didn't give you the straight dope. You were mixed up, I thought, but not how I suggested. I have copied your response here.
270.9(33.2-71.3)+4.184*264.6*
(33.2-16.9)=0

The 270.9(33.2-71.3) is correct BUT you must multiply that by the specific heat of the unknown and I don't see that anywhere. However, if I place specific heat where it should be, you appear to have solved it for the specific heat although that isn't shown. Are you copying this into a data base. If so I suspect the reason is that you reported only two significant figures but you are allowed 3. Most databases will count that wrong. Except for omission of the specific heat of the unknown all of your work is correct and the answer turns out to be 1.748 which I would round to 1.75 to 3 s.f. I looked but did not find a material with 1.75 Cp. The closest I came was for wood which gave a range of 1.7 to 2.2 or so.

I still didn't get it. Odd. I guess I'll ask my professor

To find the specific heat of the unknown compound, you need to apply the principle of heat transfer. The equation for heat transfer is:

q = m * c * ΔT

where:
q is the heat transferred,
m is the mass of the substance,
c is the specific heat capacity, and
ΔT is the change in temperature.

In this scenario, the heat gained by the water is equal to the heat lost by the unknown compound.

Let's break down the equation based on the given information:

For the water:
q1 = 264.6 g * 4.184 J/g°C * (33.2°C - 16.9°C)

For the unknown compound:
q2 = 270.9 g * c * (33.2°C - 71.3°C)

Since the calorimeter is assumed to have a negligible heat capacity, we can say that:

q1 = -q2

Therefore, we can rewrite our equation as follows:

264.6 g * 4.184 J/g°C * (33.2°C - 16.9°C) = -270.9 g * c * (33.2°C - 71.3°C)

Simplifying the equation:

-7698.3024 J = -270.9 g * c * (-38.1°C)

Now, divide both sides of the equation by (-270.9 g * -38.1°C):

c = -7698.3024 J / (-270.9 g * -38.1°C)