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
Tf
is 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 Tf
is 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.
