A155g sample of copper was heated to 150.0 degrees celsius, then placed into 250.0g water at 19.8 degrees celsius. calculate the final temperature of the mixture

Once for the water and once for the copper. Set up a table that accounts for each of the variables you know, and then identify the ones you need to obtain. Give me a moment or two and I will work this out for you.

Okay, so like I said before, you will need to use the equation twice. Now, keep in mind that when the copper is placed in the water (the hot into the cold), there is a transfer of heat. This heat transfer is measured in Joules (J). So, the energy that the water gains is the same energy that the copper loses. This means that for your two equations, they can be set equal to each other, but the copper equation will have a negative sign in front to account for the energy it's losing to the water.

When set equal to each other, the equations should resemble something like this:
(cmΔt)H20 = -(cmΔt)Cu
(Cu is copper).

Remember, Δt is the final temperature minus the initial temperature (T2-T1). We are trying to find T2. Since we are submerging the copper into the water, we can assume that the final temperature at equilibrium is the same for both the copper and the water. At a thermodynamic equilibrium, there is no heat transfer because both materials are at the same temperature.

T2Cu = T2H20

Now, the algebra for this part of the problem is a bit confusing, so make sure you keep track of your variables. If done right, the algebra should work out so you have this:

T2 = ((cmT1)Cu + (cmT1)H20) / ((cm)H20 + (cm)Cu)
Insert the values for the variables. Once you plug and chug, your final answer should be
26.8 degrees Celsius.