Energy and Human Ambitions on a Finite Planet, 2021a

A.11 Just the Start 373
way to express the answer.
It is also useful to pause <strong>and</strong> reflect on our operations <strong>and</strong> whether
they made sense. For instance, since we multiplied by mass to get the
cooling time, it implies that a larger brick would take longer, which
is sensible. If heat left at a rate faster than 10 W, it would cool down
faster, again making sense.
We will do one more example in an unfamiliar context, this time involving
some ambiguity that your wits can help resolve.
Example A.10.5 An outer wall on a sealed brick building measures
5 m long <strong>and</strong> 2.5 m high, having a thickness of 0.1 m. You are asked
how fast heat (thermal energy) is being lost through the wall. 37 It is
20 ◦ C warmer inside the wall than it is outside, <strong>and</strong> you are told that
brick has a thermal conductivity of 0.6 W/m/ ◦ C. 38 Thermal what?
But don’t panic.
At this point, we could create our own formula
based on requiring the units to work!
See: formulas are not sacred tablets to be
memorized—they are just statements that
make logical sense <strong>and</strong> can be created by
you to accomplish a task.
37: The units would be energy per time, or
J/s, which is a power (W).
38: This construction means that m <strong>and</strong> ◦ C
are both in the denominator together.
We want an answer in J/s, or W. We see Watts in the numerator of
the thermal conductivity, so we want that in the numerator of our
answer. We would need to multiply by meters <strong>and</strong> by ◦ C to cross the
finish line. Seems simple enough. But meters shows up three times
in the dimensions of the wall. Which should we choose? Or is it a
combination?
Engage the intuition. Put yourself in the building next to the wall,
mentally. It’s warm inside, cold outside. Will I need more heaters
(power) or fewer if the wall is taller? If the wall is wider? If the wall is
thicker? What does your intuition say?
You might reason that a thicker wall will require less power to keep
warm, but that a larger area (increasing width or height or both)
would make the job of maintaining temperature more challenging.
This suggests that the power will increase if width or height increase,
<strong>and</strong> decrease if thickness increases. So we should multiply by width <strong>and</strong>
height (or equivalently, by area) <strong>and</strong> divide by thickness. Area over
thickness indeed has units of meters, which we already concluded
we needed as a multiplier to get our desired outcome. Putting things
together,wehave
W
0.6
m ·◦ C
· 5m· 2.5m
0.1m · 20◦ C 1, 500 W,
which is about the output of one space heater.
A.11 Just the Start
It is well beyond the scope of this book to engage in an exhaustive
review of math concepts. Hopefully what has been covered provides a
© 2021 T. W. Murphy, Jr.; Creative Commons Attribution-NonCommercial 4.0 International Lic.;
Freely available at: https://escholarship.org/uc/energy_ambitions.