A 25.78g block of Fe(s) at 47.15℃ is dropped into 100.0g of ethanol at 4.87℃. When the system reached thermal equilibrium, what is the temperature? Assume the system is perfectly insulated/isolated. Heat capacity of Fe(s) = 0.451J/g℃and of ethanol = 2.46J/g℃.
This is a “coupled systems” problem. The “hot” Fe(s) will lose energy/heat that will be absorbed by the “cold” ethanol. Qualitatively, the final temperature must be less than 47.15℃ and greater than 4.87℃, given the relative amounts and properties of Fe(s) and ethanol, it's reasonable to assume that the final temperature will be closer the 4.87℃ than 47.15℃. Because the system is “perfectly insulated”, all of the energy lost by the Fe(s) will be gained by the ethanol. So the energy picture is:
Eout(from Fe(s)) = (0.451J/g℃)(25.78g)(Tf – 47.15℃)
Ein(to ethanol) = (2.46J/g℃)(100.0g)(Tf – 4.87℃)
Let's look at 2 ways to treat this, one is more rigorously mathematical, the other in still mathematical, but involves a little hand-waving. You can decide which is which...
Using the equations as written above, we have a “frame of reference” problem. Eout should equal Ein, but in order to do the straight-up algebra, we need to approach the problem from a single frame of reference. What does this mean? The short answer? Sneak a little negative sign into the equation. The effect of this will be to change the frame of reference of one of these energies so that they're both the same. Then it's an algebra problem to solve for Tf:
- (0.451J/g℃)(25.78g)(Tf– 47.15℃) = (2.46J/g℃)(100.0g)(Tf– 4.87℃)
For simplicity's sake, I know that all of the units are going to cancel (except for the final ℃), so I'm going to drop them. Let's bunch all of the constants together...
(-0.04726)(Tf – 47.15℃) = (Tf – 4.87℃)
Distributing the constant...
-0.04726Tf + 2.22847 = Tf – 4.87
Grouping “Tf” terms and number terms...
2.22847 + 4.87 = (1 + 0.04726)Tf
7.098 = 1.04726Tf
And the final step gives...
Tf = 6.78℃
If we'd rather treat this as a “magnitude of energy” problem, then we don't have to mess around with adding a negative sign. We can do this by using the absolute value of the change in temperature. Do you have an “absolute value” button on your calculator? I don't think I do... Fortunately, this is a real-world problem, so we can use a little intuition to make thing behave. Since Tfis greater than 4.87, the (Tf– 4.87℃) term will be positive, so the absolute value takes care of itself. The other temperature change, (Tf– 47.15℃), will be negative {since we know that Tfis less than 47.15}, so the absolute value of this term will be... (47.15℃ – Tf). Now we can plug in and once again solve:
(0.451J/g℃)(25.78g)(47.15℃ – Tf) = (2.46J/g℃)(100.0g)(Tf– 4.87℃)
(0.04726)(47.15℃ – Tf) = (Tf – 4.87℃)
2.22847 – 0.04726Tf = Tf – 4.87℃
2.22847 + 4.87 = (1 + 0.04726)Tf
7.098 = 1.04726Tf
Tf = 6.78℃
Same result either way.
