Specific Heat Problem, not getting the correct answer

[HRubss]

Homework Statement

A copper pot with a mass of 0.500 kg contains 0.170 kg
of water, and both are at a temperature of 20.0°C. A 0.250-kg
block of iron at 85.0°C is dropped into the pot. Find the final temperature
of the system, assuming no heat loss to the surroundings.

Homework Equations

Q = cmΔT
ΣQ = 0

The Attempt at a Solution

Since I know that all these materials will reach thermal equilibrium and that the iron will be the only one losing heat so...

C_copperm_copperΔT + C_waterm_waterΔT - C_ironm_ironΔT = 0

I expanded all the specific heats, brought the individual heats at initial temperature and then solved for T_final. I got...
T_final = (C_copperm_copperT_initial+C_waterm_waterT_initial-C_ironm_ironT_initial) /(C_copperm_copper+C_waterm_water-C_ironm_iron)

I converted to Kelvin and got about 10.47°C, the answer should be 27.45°C... I don't understand what I did wrong.

 

[kuruman]

C_copperm_copperΔT + C_waterm_waterΔT - C_ironm_ironΔT = 0

You should put a plus sign in front of C_ironm_ironΔT and you shouldn't really use the same symbol ΔT to denote two temperature changes in this case. The equation says that the sum of all the heat transfers must be zero as the iron loses heat (negative ΔT_iron) and the copper and water gain heat (positive ΔT_copper+water). In other words, the changes in temperature are algebraic quantities, that could be positive or negative.

 

[HRubss]

You should put a plus sign in front of C_ironm_ironΔT and you shouldn't really use the same symbol ΔT to denote two temperature changes in this case. The equation says that the sum of all the heat transfers must be zero as the iron loses heat (negative ΔT_iron) and the copper and water gain heat (positive ΔT_copper+water). In other words, the changes in temperature are algebraic quantities, that could be positive or negative.

So the equation should be like this?
ΔQ(copper) + ΔQ(water) = - ΔQ(iron)

 

[gneill]

So the equation should be like this?
ΔQ(copper) + ΔQ(water) = - ΔQ(iron)

Sure, or equivalently:

ΔQ(copper) + ΔQ(water) + ΔQ(iron) = 0

The trick is to get the ΔQ's correct for each term, and that means paying attention to the signs of the temperate changes that apply to each of the ΔQ's (some materials are rising in temperature, gaining heat, while others are lowering in temperature, losing heat).

 

[HRubss]

Sure, or equivalently:

ΔQ(copper) + ΔQ(water) + ΔQ(iron) = 0

The trick is to get the ΔQ's correct for each term, and that means paying attention to the signs of the temperate changes that apply to each of the ΔQ's (some materials are rising in temperature, gaining heat, while others are lowering in temperature, losing heat).

So...
Substituting the entire formula:
CMΔT(copper) + CMΔT(water) = -CMΔT(iron)
I know that ΔT of copper and water should be the same and I want to get all the T(final) all on one side:
ΔT(CM(copper) + CM(water)) = -CMT(final, iron) + CMT(initial, iron),
and from here, I split up the ΔT of copper/water and then solve for T final by getting the iron's Tfinal on the other side?

 

[gneill]

Sure, that's the idea.