Physics Group 4
Project ??????????? After being
set our initial topic of the bath tub, we chose as a group to investigate the
heat lost from a bath tub.? We initially
experimented with ideas such as using different shapes, sizes and materials.? Eventually, we decided that the easiest
option was to investigate how the thickness of a bath tub effects the heat
loss, as well as how conduction, convection (evaporation) and radiation
effected the heat lost from a bath tub and how we could reduce these forms of
heat loss. We chose to use vacuum formed plastic bath tubs as they were the
quickest to build, and we did not want to waste valuble experimenting time
building lots of bath tubs.? The
polyethene plastic also mimicked the material of a real bath tub fairly
successfully.The investigation of energy (heat) lost in different bath
tubs. ??????????? We began
our investigation by determining the specific heat capacity of water. In order
to do this, we took a known mass of water in an insulated beaker and gave it a
known quantity of energy by using an emersion heater.? We also used an accurate thermometer in order to take the
temperature change in the water.? We
then used the equation: ??????????? E = mc∆t We then rearranged the formula to give: ??????????? c = E/m∆t In order to calculate the specific heat capacity of water
(c), we took the known quantities of mass, energy and the change in temperature
and inserted them into the above equation. In order to calculate the energy
given to the water, we used electricity to heat the water, by doing this we
were able to calculate the energy using P = VI.? We measured both the voltage and the current that we used to heat
the water.? We had previously weighed
the water and hence we knew the mass, and the change in temperature we measured
using a thermometer.? Hence we were able
to calculate the following values for the specific heat capacity of water: Test 1 Test 2 Test 3 Average Specific heat capacity of water 4745 4723 4216 4561 ??????????? The actual
specific heat capacity of water is 4200.?
We would expect our results to be lower than this value.? However, the beaker was very well insulated
with a lid and hence little heat would have been lost to the surroundings.? The high values can be explained due to: –
The inaccuracy of the probe thermometer we used.? We later found it was not the most accurate
thermometer available to us.? For the
remainder of our experiments we used a far better digital probe type
thermometer. –
The inaccuracy of the digital ammeters, voltmeters and the
unreliability of the power pack and the immersion heater. –
The impurities in the water which was not deionised.Background to our experiment ??????????? We
researched the different methods of heat loss and came up with the following
information: Conduction: The mechanism by which heat energy is
transferred through solid materials. In the case of our experiment, the heat
from the water will be transferred to a certain extent through the plastic
walls of the bath tub by conduction. Evaporation: The mechanism by which most energetic molecules
escape from the water surface leaving the less energetic molecules behind.? This means the internal energy of the water
is then lowered and its temperature is decreased. Convection: The mechanism by which fluid ?currents?, caused
by the differential densities of hot and cold fluids, move.? As the fluid gets ?hot? the particles
vibrate more, and therefore expand, become less dense and rise.? These particles then cool, become less dense
and sink. For the purpose of this experiment, I will consider the heat
lost by evaporation and conduction as one and the same thing as they are nearly
impossible to separate in a liquid such as water. Radiation: The mechanism by which energy from the water is
lost to the surroundings by electromagnetic waves.? This occurs even at low temperatures.? The Investigation ??????????? We began
our experiment by building two vacuum formed bath tubs of different
thicknesses.? We decided that in order
to investigate the effect of thickness as well as the heat loss from
conduction, convection and radiation we would undertake the following
experiments. 1)
We would fill both bath tubs with 700g of water and heat them
up to approximately 42.5 degrees, the temperature of an average bath tub.? We would then measure the heat lost from
both the thick and thin bath tubs in a 45 minute period.? We will record the change in temperature
every five minutes and from this hopefully plot a heat loss curve. 2)
We will then put a lid on the bath tubs to reduce the heat
loss due to evaporation and repeat the same experiment. 3)
We will then put insulation around the bath tubs to reduce the
heat loss due to conduction and repeat the same experiment. 4)
We will the paint the bath tubs silver to reduce the heat loss
due to radiation and repeat the same experiment. From these experiments we hope to
be able to calculate the heat loss due to conduction, convection and radiation
by virtually eliminating each of these forms of heat loss in turn and thereby
calculating? how much is lost through
each of these forms of heat loss. ????? We
wanted to make our measurements of heat loss as accurate as possible.? After our initial experience with the
inaccuracy of the probe we used for our heat capacity of water experiment we
wanted to have a more accurate thermometer. We experimented with a thermocouple
but the galvanometer was too sensitive for our purposes.? We therefore used a more accurate digital
probe and calibrated it before each experiment in order to calculate the error
in our readings.Experiment 1: ????? In
this experiment I used two vacuum formed bath tubs of differing thicknesses. Results: Time
(mins) Heat Lost
(’C) Heat Lost
(’C) 5 1.6 2.7 10 1.5 1.7 15 1.2 1.3 20 1.1 1.1 25 1 1.3 30 0.8 0.7 35 0.8 0.8 40 0.7 0.7 45 0.7 0.6 Total
Heat Lost 9.4 10.9 I have drawn two graphs of these
results to show the heat loss curve by plotting the heat lost against time for
both the thick and thin bath tubs.Experiment 2: ????? In
this experiment I used two vacuum formed bath tubs of differing thicknesses
with lids on them. Results: Time
(mins) Heat Lost
(’C) Heat Lost
(’C) 5 2 2.3 10 1.2 1.2 15 1 0.9 20 0.8 0.8 25 0.7 0.8 30 0.6 0.6 35 0.6 0.6 40 0.6 0.5 45 0.5 0.4 Total
Heat Lost 8 8.1 Experiment 3: In this
experiment I used two vacuum formed bath tubs of differing thicknesses
surrounded by insulation. Results: Time
(mins) Heat Lost
(’C) Heat Lost
(’C) 5 1.7 1.9 10 1.2 1.3 15 1 1.1 20 0.9 1 25 0.8 0.9 30 0.8 0.9 35 0.7 0.9 40 0.6 0.8 45 0.2 0.7 Total
Heat Lost 7.9 9.5 Experiment 4: In this experiment
I used two vacuum formed bath tubs of differing thicknesses painted silver. Results: Time
(mins) Heat Lost
(’C) Heat Lost
(’C) 5 2.2 2.3 10 1.3 1.5 15 1.4 1.4 20 1.1 1.3 25 1 1.1 30 0.8 0.9 35 0.8 0.9 40 0.8 0.8 45 0.5 0.7 Total
Heat Lost 9.9 10.9 Graph of the Total Heat
LostConclusions to be drawn
from our graph: ·
The lid and the surrounding insulation had a
significant impact on the heat lost from the bath tub. ·
Painting the bath tubs silver had no real impact ·
The thick bath tub lost more heat than the thin one. ·
The silver painted bath tub lost more heat than the
normal bath tub. ·
Heat loss was not uniform but varied.Conclusion: ??????????? As the
insulation and lid on the bath tub made a significant difference to the heat
lost from the bath tub, we assumed initially that the heat lost through
conduction and evaporation had been significantly reduced by the insulation and
the lid respectively.? However, on
further investigation we found that plastic is such a poor conductor that it is
unlikely that any heat was lost was lost through this method and it is more
likely that the heat that we initially assumed was lost through conduction was
lost through radiation.? We can
therefore assume that the greatest amount of heat is lost through evaporation
and some was lost through radiation, in comparison the heat lost through
conduction was minute.? The thickness of
the bath tub also makes a difference, as the thicker the bath tub is, the greater
the insulation, the greater the reduction of radiation.? The silver painting did not make much
difference as although some heat was reflected by it, the bath tubs were so
thin that most of the heat escaped and in comparison, the heat reflected back
by the silver was almost immeasurable. ??????????? The results
of our experiment showed conclusively that most of the heat was lost through
evaporation, radiation also had a significant effect as when the bath was
insulated the heat loss was reduced significantly. We can therefore conclude
that the best bath tub would be one that was insulated and with a lid on it.Evaluation: ??????????? Our results
were, on the whole, fairly accurate, however, we did encounter some
difficulties.? Firstly, the temperature
probes we used were extremely unreliable, despite calibrating them for errors,
it would take at least 30 seconds for them to settle down and reach the right
temperature and therefore it is likely that during the 30 seconds the
temperature would have dropped considerably and hence making our results
slightly in accurate.? In order to
irradicate this, if I was to repeat the experiment I would use a computerised
temperature measuring device. This would enable me to accurately plot a graph
as the computer could be programmed to take the results at far more regular
intervals far more accurately.?
Secondly, our methods of eliminating heat loss were not entirely
reliable.? In each case it is likely
that despite our best efforts to reduce heat loss, our insulation was not 100%
reliable.? The worst of all our attempts
to reduce heat loss was the bath tub we painted silver as the silver paint
actually melted the thin bath tub making the plastic thinner and thus
increasing the heat loss, hence the inconsistencies in the graph above.? If repeating the experiment I would be
careful to use thicker plastic and less silver paint.? Furthermore, despite good results, I felt the 1mm difference in
the thickness of the two bath tubs was insufficient to give a real indication
of the effect of thickness on heat loss.?
In a future experiment, I would use far thicker plastic to differentiate
between the two.? Other problems were
encountered with our poor fitting insulated lids which could have been
improved. ??????????? Despite our
problems, I feel that our experiment has been a success, we worked together well
as a group and our results on the whole were valid.? Although our experiment could have been improved, and in
conducting a future experiment, I would make amendments to the temperature
measuring equipment particularly, through the experiment, I feel our
experimental technique improved and the experience of doing a longer experiment
was extremely useful.
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