HOME
HEATING
Heat
Loss
or
Gain
of
an
Insulated
Container
Take‐Home
Experiment
Write
Up
Task
1a:
Measuring
and
Modelling
Heat
Loss
of
an
Insulated
Container
According
to
Newton’s
Law
of
cooling,
the
rate
of
heat
loss
from
an
object
should
be
proportional
to
the
difference
in
temperature
between
that
object
and
its
surroundings.
Heat
is
transferred
in
three
ways:
conduction,
convection
and
radiation.
Conduction,
the
most
relevant
to
our
experiment,
is
the
transfer
of
heat
from
high
temperature
to
low
temperature
by
the
direct
contact
of
atoms
or
molecules.
Convection
is
the
transfer
of
heat
by
gases
or
liquid
in
motion
and
radiation
is
the
transfer
of
heat
through
space
(for
example
heating
of
Earth
by
the
Sun).
Thus
to
model
heat
transfer
of
this
container,
measured
heat
loss:
ΔQ mcΔT
=
Δt
Δt
will
be
compared
to
predicted
heat
loss
due
to
conduction:
€
ΔQ kA(T − T0 )
=
Δt
d
To
measure
heat
loss,
a
Styrofoam
container
filled
with
water
was
heated
in
a
microwave
(about
one
€
minute
on
high).
The
container
was
made
from
one
Styrofoam
cup
and
one
Styrofoam
bowl,
purely
out
of
the
convenience
of
using
readily
available
materials
(See
Figure
1).
Figure
1:
The
insulated
container
was
made
from
a
Styrofoam
cup
and
bowl
glued
together.
An
oven
thermometer
measured
the
temperature
of
the
water
through
a
hole
in
the
top.
Physics
and
Astronomy
Outreach
Program
at
the
University
of
British
Columbia
The
cup
was
filled
with
water
and
the
bowl
was
glued
to
its
top
with
epoxy
glue.
The
cup
should
be
filled
very
full
so
that
the
water
touches
the
top
(the
bowl)
but
does
not
leak
out
of
the
hole
in
the
top.
A
marble
was
also
added
to
enable
stirring.
A
hole
was
drilled
into
the
top
of
the
container
so
that
a
digital
oven
thermometer
could
measure
the
temperature
of
the
water.
The
surface
temperature
of
the
container
was
measured
with
an
infrared
thermometer.
It
was
found
that
the
top,
bottom
and
sides
were
at
slightly
different
temperatures,
so
each
was
recorded.
The
hot
water
was
then
left
to
cool
to
room
temperature
and
the
temperature
of
the
water
and
the
surfaces
of
the
container,
were
recorded
at
different
time
intervals
–
once
a
minute
to
start
with,
and
every
5
or
10
minutes
later,
as
the
rate
of
cooling
slowed.
The
container
was
held
on
a
ring
stand
so
that
most
of
the
Styrofoam
was
in
contact
with
air.
Before
every
measurement
the
container
was
stirred
or
shaken
for
at
least
10
s
until
the
water
temperature
stabilized.
Data
Analysis
and
Interpretation
Plot
theoretical
(predicted)
heat
flow
and
experimental
heat
flow
against
time
to
explore
how
well
losses
due
to
conduction
account
for
the
actual
energy
loss.
The
constants
known:
Conductivity
of
Styrofoam:
k
=
0.033
W/m/K
Specific
Heat
Capacity
of
Water:
c
=
4200
J/kg/K
The
variables
measured:
Mass
of
water:
m
=
0.2232
±
0.0001
g
Thickness
of
Styrofoam:
dtop
=
0.0030
±
0.0002
m,
dbottom=0.0019±0.0002
m,
dsides=0.0017±0.0002
m
Radius
of
bottom
(rbottom)=
0.025
±
0.001
m
Radius
of
top
(rtop)=
0.040
±
0.001
m
Slant
height
(s)
=
0.090
±
0.001
m
Surface
area:
A
=
?
Calculating
surface
area:
For
this
experiment,
we
will
take
the
surface
area
to
be
that
of
one
Styrofoam
cup
with
the
top
covered
as
this
was
the
only
part
of
the
Styrofoam
actually
in
contact
with
the
water.
To
calculate
the
area
of
the
side
of
the
cup,
use
the
formula
for
the
surface
area
of
a
right
circular
cone.
r
bottom
A = Area of thebottom + Area of the top+ Area of the side
2
2
= πrbottom + πrtop + π (rbottom + rtop )s
= π (.025 m) 2 + π (.04 m) 2 + π (0.025 m + 0.04 m)(0.09 m) 2
2
= 0.002 m + 0.005 m + 0.0183 m
= 0.0253 m 2
€
Physics
and
Astronomy
Outreach
Program
at
the
University
of
British
Columbia
slant
height
(s)
2
r
top
The
heat
loss
due
to
conduction
for
a
temperature
difference
can
be
predicted
using
the
formula:
where
ΔT
is
the
temperature
difference
between
the
water
and
its
surroundings.
In
this
case,
its
surroundings
were
the
top,
bottom
and
sides
of
the
container
which
each
had
slightly
different
temperatures
at
any
given
time.
Thus
the
total
energy
loss
due
to
conduction
can
be
predicted
to
be
(at
any
given
time):
These
predicted
energy
losses
can
be
plotted
against
time
to
get
a
theoretical
cooling
curve.
This
curve
can
be
compared
to
the
actual
measured
energy
losses,
calculated
using
measured
temperature
differences
and
times
using:
The
change
in
temperature
at
a
time
point
was
calculated
by
finding
the
difference
between
the
temperature
at
the
next
time
point
and
the
previous
time
point.
For
example
at
t=2
min,
ΔT=T3min
–
T1min,
and
Δt
=
3
min
–
1
min
=
2
min.
Figure
2
shows
both
curves.
Vertical
error
bars
are
included
for
P(experimental).
Figure
2:
Plots
of
theoretical
and
experimental
heat
loss
over
time.
Physics
and
Astronomy
Outreach
Program
at
the
University
of
British
Columbia
If
we
have
correctly
modelled
the
sources
of
heat
loss
we
expect
that
the
two
curves
should
overlap,
which
they
do.
It
is
also
possible
to
use
the
data
(a
spread
sheet
of
sample
data
is
available
for
this
activity)
to
determine
the
conductivity
of
the
container.
If
the
conductivity
were
unknown
the
value
for
k
(in
the
theoretical
energy
loss
calculation)
could
be
altered
to
shift
the
theoretical
plot
onto
the
experimental
plot.
However,
in
this
case,
the
plots
align
nicely
with
the
accepted
value
for
k
=
0.033
W/m/K.
Thus
we
have
shown
that
the
heat
loss
can
be
successfully
modelled
taking
only
conduction
into
account
and
that
losses
due
to
radiation
and
convection
are
negligible.
Some
procedural
specifications
to
take
note
of:
• Thicknesses
were
measured
with
a
caliper.
• Care
was
taken
to
make
sure
very
little
water
escaped
from
the
hole
in
the
top.
It
if
did
it
was
dried
up
with
a
napkin
immediately
because
if
the
top
of
the
container
were
wet,
it
would
have
significantly
different
heat
transfer
due
to
the
evaporation.
• Measurement
errors
were
mostly
ignored,
however
to
create
the
error
bars
in
Figure
1,
errors
of
0.008
min
and
0.5°
were
used.
• The
specific
heat
capacity
of
the
marble
was
ignored.
• It
was
difficult
to
obtain
simultaneous
measurements
of
the
temperature
of
the
sides,
top
and
bottom
of
the
container.
They
were
always
measured
in
the
same
order
but
there
is
some
systematic
error
in
the
time‐temperature
correlation.
Summary
Modelling
the
heat
loss
with
conduction
took
into
account
most
of
the
energy
lost
by
the
water
so
that
the
theoretical
and
experimental
heat
losses
against
time
produced
similar
curves.
Therefore
most
of
the
heat
was
lost
due
to
conduction
between
water
and
the
Styrofoam
cup.
Brittany
Tymos
and
Rachel
Moll
2009/08/27
Physics
and
Astronomy
Outreach
Program
at
the
University
of
British
Columbia