PASCO ET-8782 Energy Transfer– Thermoelectric - Manual (Page 16)

Energy Transfer –Thermoelectric Teachers’ Notes–Conservation of Energy
16 ®
5) Most of the heat that flowed out of the two-reservoir system was lost. Some of it flowed
through the foam insulators to the environment. Some of it was dissipated in other parts of the
circuit.
6)
7) W
observed
= Q
hot
– Q
cold
– E
lost
8) The result of doing work on the resistor was that the resistor dissipated heat to the
environment. For a more practical use of the useful work, the resistor could have been
replaced with a light bulb, an electric motor, or some other electrical device.
Conservation of Energy
9)
10) This is not a good way to store energy.
Conduction and Heat Flow Through Insulator
11)
Q
i
= (0.031 J/s) × (150 s) = 4.6 J
12) Q
i
is small compared to Q
cold
.
13) This is an estimate of the heat that flowed from the outside air, through the insulator, and into
the front face of the aluminum block on the cold side. Some more heat flowed in through the
sides that we ignored. Q
i
is likely an overestimate because the actual temperature difference
was not always as large as the
∆T that was used in the calculation, and the surface area of the
front face is larger than that of the sides.
14) Heat flow through the insulator on the hot side would be larger in magnitude because there
was a greater temperature difference between the block and the outside air. Since the block
was hotter than the air, heat would have flowed out to the environment.
15) The amounts of “lost energy” in the Heat Pump and Heat Engine phases were 12.5 J and 2.7 J.
The estimate of Q
i
suggests that heat flow through the insulators was a significant
contribution to this unaccounted-for energy. Another possible contribution to the lost energy
is heat dissipated by other components of the circuit, especially the material inside the peltier.
16) Without the insulators, it is likely that the net heat flow to the environment would have been
greater, thus increasing the amount of lost energy.
% of useful work
0.572 J
( )
3.3 J
( )
---------------------- 100 %
×
17 % = =
% recovered
0.572 J
60.3 J
----------------- 100 %
×
0.9 % = =
Q
i
t
⁄
0.36 W/(m·°C)
[ ]
0.033 m
( )
0.037 m
( ) × [ ]
7 °C
( )
0.01 m
( )
--------------------- Ч Ч
0.031 J/s = =