Daniel Alvarez
Maria Winters
Jordan Wisner
Jeffery Jordan
ENGIN 101
Transfer of Heat between a Frozen Water Bottle, Water Bottle, and the
Environment
Abstract
This lab looked into the heat transfer between a frozen water bottle and a room
temperature water bottle. The overall goal was to cool the warm water bottle as much as
possible. Our design consisted of a fiberglass isolator inside with a thin strip of aluminum
foil in between the two bottles in contact length-wise. We had an hour to cool the bottle.
Starting at around 15 degrees Celsius, we cooled the bottle to 8.825 degrees Celsius.
55,094 joules of energy went into the frozen water bottle or into the enviorment.
Introduction
This report discusses a project designed to maximize heat loss between two bottles of
water under certain conditions. The importance of this experiment was to understand the
different methods for cooling water under certain conditions. It was important to
understand the different forms of heat transfer and how to calculate predicted
temperatures. The conditions applying to the experiment involved one frozen water
bottle at 0°C and another water bottle at room temperature. The water bottles were
not allowed to be opened during the one hour interval of the experiment. A $15.00
budget was allowed to purchase whatever items thought to be the most useful in the
experiment. The objective of this experiment was to, in one hour, use the frozen water
bottle to cool the room temperature water bottle to the lowest possible temperature.
Using the $15.00 budget, the items purchased to achieve lower temperature were
aluminum foil and fiber glass insulation. The reason for aluminum foil was because of its
excellent heating conductivity. A layer of aluminum foul was placed in between the two
water bottles. This allowed for strong heat exchange. Fiber glass insulation was then
wrapped around the two water bottles to keep the heat from escaping into the atmosphere.
The foil allowed the room temperature water bottle to cool, while the insulation kept it
cool. This method was very cheap, efficient and easy.
Procedure
After examining the bottles we would be using in a cooling system we began
brainstorming ideas on how to incorporate a frozen water bottle into a system that could
cool another bottle containing room temperature water. One of our first ideas was to cool
down the room temperature bottle by using a pump to circulate water between the bottles
and cool via convection. We decided not to use this idea on the basis that we could not
come up with any pumps that could be assembled from household materials without
Daniel Alvarez
Maria Winters
Jordan Wisner
Jeffery Jordan
ENGIN 101
using a power supply.
In interest of using household materials we briefly considered using evaporating alcohol
to cool down a water bottle but weren’t able to incorporate the frozen water bottle and
decided to use a design that only used conduction in order to model heat transfer more
easily. Our original heat transfer system consisted of wrapping the two water bottles in
aluminum foil and surrounding the system with fiberglass insulation to reduce heat
transfer from the air into the system. After initially testing this design we decided to
simplify the system further by placing a single rectangular piece of aluminum foil
between the water bottles rather wrapping the bottles entirely in aluminum foil.
\RESULTS/DISCUSSION
In a time period of one hour, or 3600 seconds, the final temperature of the originally
room temperature water bottle was 15oC. The frozen water bottle had not completely
melted as well. The room temperature water bottle had lost 15.303964 J/s (Watts) of heat,
for a total of 55,094 Joules to the frozen water bottle. The room temperature water bottle
also lost 6.479304 J/s (Watts) of heat, for a total of 23,325.4944 Joules to the surrounding
room atmosphere through the insulation. These two values were calculated using the
equation
q= A(Tins-Tout) / ΣR. The total amount of heat lost from the warm water bottle was As a
result of calculations, the final temperature predicted was 8.825oC using the equation E=
mcΔT Thus, the percent error was 41.1% between the predicted final temperature and the
actual final temperature. For equations and calculations used, see Appendices A and B.
Overall, our design ended up working in cooling down the room temperature water
bottle. We did overestimate how much the water bottle would cool down, but this could
have been due to several factors. For one, our design was held together with staples,
meaning that heat could have escaped through the gaps between the staples. This would
have meant a possible error in our calculations. Another source of error was that it was
difficult to estimate the contact area between the two water bottles, due to the fact that it
was just difficult to see or measure. So, this value was estimated based on the dimensions
of the foil strip that was in between the two water bottles by using a value that was
approximately an eighth of the area of the strip. This value was crucial in calculating the
heat loss from the room temperature water bottle to the frozen water bottle in our device.
Another error was estimating the temperature of the room temperature water bottle
compared to the real room temperature of the environment. We estimated it to be 26oC,
which is little warmer than the room temperature of 24oC because water has a higher
thermal conductivity than air. So, this potentially could have been an error in our
experiment as well, because we assumed that that 26oC was the entire inside temperature
inside our device. To improve our design, we could have wrapped both of the water
bottles completely in aluminum foil, as well as increased the amount of foil in between
Daniel Alvarez
Maria Winters
Jordan Wisner
Jeffery Jordan
ENGIN 101
the two bottles. This would have increased the contact area between the water bottles, and
thus could have sped up the heat loss from the warm water bottle to the frozen one. We
could also potentially sew the edges of our device together instead of stapling, thus
minimizing the chances of heat loss or heat gain to/from the outside environment. We
also could have taken the initial temperature of the room temperature water bottle. If we
had had a better understanding of the equations and concepts, our design could have been
better from the start, and we possibly would have been able to calculate the final
temperature more accurately of the initially-warm water bottle.
Conclusion
In conclusion, there was a large amount of elements which we could not control. The
openings in our insulation, the contact between the two water bottles, the contact between
the aluminum foil and two bottles. We overestimated the contact area between the
aluminum foil and the water bottles assuming that the foil completely contacted between
the water bottles. This was not the case. It only contacted at small percentage of the area
we thought it did which drastically affected out results.
Appendix A
Heat transferred from warm water bottle to frozen water bottle:
q= A(Tins-Tout)/ ΣR
A= 1.288x10-4m2
Tins= 24oC
Touts=0oC
ΣR= Rplastic +Rfoil+Rplastic = .00250085 m2*K/W
where
ΣR = Lplastic + Lfoil + Lplastic
Kplastic Kfoil
Kplastic
Kplastic was K of polyethylene terephthalate which is .4 W/m*K
Lplastic was estimated to be .5mm= 5x10-4m
Kfoil was 235 W/m*K
Lfoil is .2mm= 2x10-4m
Thus, q= -15.303964 (negative because room temperature water bottle lost heat)
Daniel Alvarez
Maria Winters
Jordan Wisner
Jeffery Jordan
ENGIN 101
Appendix B
Heat transferred from room temperature water bottle to outside environment through the
insulation:
q= A(Tins-Tout)/ΣR
Tins=26oC
Touts=24oC
6 different A’s to represent the six normal areas to heat flow of the device
A1= .024m2
A2=.00742m2
A3= .00742m2
A4= .0312m2
A5= .0636m2
A6= .0636m2
ΣR= Rplastic +Rinsulation +Rair
ΣR = Lplastic + Linsulation + Lair
Kplastic Kinsulation Kair
Lplastic= .5mm=.5x10-3m
Kplastic= .4 W/m*K
Rinsulation= 6.7 m2*K/W
Lair= .0675m for A2/A3 and .0435m for A1 and A4
Kair= .024 W/m*K
Calculating q 6 times, one for each area A, gives a total of -6.479304 J/s (W) lost from the room
temperature water bottle to the surrounding environment.
Appendix C
To calculate the final temperature of the initially room temperature water bottle we calculated the
total heat lost from the room temperature water bottle:
qtotal= qinside-qoutside
= -15.303964 W – (-6.479304 W)
= -8.82466 W = -8.82466 J/s
Daniel Alvarez
Maria Winters
Jordan Wisner
Jeffery Jordan
Energy = Power * time
E= -8.82466 J/s * (3600s)
= -31.768776 kJ
E= mcΔT
E = mc (Tfinal- Tinitial)
Solving for Tfinal yields:
Tfinal = (E/mc) + Tinitial
Tfinal = (-31.768776kJ/[(.5kg)*(4.187kJ/kg*K)]) + 297.15K
= 281.975K
Tfinal =8.825oC
ENGIN 101