3B Scientific Heat Equivalent Apparatus - Manual (Page 7)

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R R R R R / k / k / k / k / k Ω Ω Ω Ω Ω T T T T T / / / / / °C C C C C R R R R R / k / k / k / k / k Ω Ω Ω Ω Ω T T T T T / / / / / °C C C C C R R R R R / k / k / k / k / k Ω Ω Ω Ω Ω T T T T T / / / / / °C C C C C R R R R R / k / k / k / k / k Ω Ω Ω Ω Ω T T T T T / / / / / °C C C C C R R R R R / k / k / k / k / k Ω Ω Ω Ω Ω T T T T T / / / / / °C C C C C
7.86 14.97 6.78 18.19 5.70 22.05 4.62 26.84 3.54 33.10
7.84 15.03 6.76 18.26 5.68 22.13 4.60 26.94 3.52 33.24
7.82 15.08 6.74 18.32 5.66 22.21 4.58 27.04 3.50 33.38
7.80 15.14 6.72 18.39 5.64 22.29 4.56 27.14 3.48 33.51
7.78 15.19 6.70 18.45 5.62 22.37 4.54 27.24 3.46 33.65
7.76 15.25 6.68 18.52 5.60 22.45 4.52 27.35 3.44 33.79
7.74 15.31 6.66 18.58 5.58 22.53 4.50 27.45 3.42 33.93
7.72 15.36 6.64 18.65 5.56 22.61 4.48 27.55 3.40 34.07
7.70 15.42 6.62 18.72 5.54 22.69 4.46 27.66 3.38 34.22
7.68 15.47 6.60 18.78 5.52 22.77 4.44 27.76 3.36 34.36
Relationship between resistance and temperature of the temperature sensor
temperature distribution, the resistance of the
temperature sensor is R
1
= 8.00 k
Ω
(correspond-
ing to T
1
= 14.60°C by Eq. 1).
•
After zeroing the counter, the experiment is be-
gun by turning the crank and thus lifting the pri-
mary weight from the ground. This slightly loos-
ens the cord so that it causes less friction on the
cylinder. The primary weight no remains at the
same height and should remain there for the rest
of the experiment.
•
After n = 460 turns the experiment is halted and
the resistance value read off: R
2
= 3.99 k
Ω
(T
2
= 30.26 °C). Since the temperature continues
to rise for a short time after the experiment is
completed (homogenizing the temperature distri-
bution), the minimum value of the resistance is
noted as the measured value. This is reached a
few seconds after the end of the experiment. Af-
ter that the resistance increases again since heat
is exchanged with the environment to cool the
cylinder down to a lower temperature.
4.1.2 Experiment evaluation
•
Work W is defined as the product of force F and
displacement s
W = Fs
(2)
•
The force of friction acting is
F = m
A
g (3)
(g is the acceleration due to gravity) in the direc-
tion of the displacement
s = n
π
D
r
(4)
•
Placing Equations 3 and 4 into Equation 2 gives:
W = m
A
gn
π
D
R
= 5.22 x 9.81
x 460 x 3.1416 x 0.04575 Nm = 3386 Nm (5)
•
The heat stored in the friction cylinder
∆
Q is
determined from the temperature difference
(T
2
– T
1
) and the specific heat capacity given in Sec-
tion 2:
∆
Q = c
A
m
A
(T
2
– T
1
) = 0.86 x 0.249
x (30.26 – 14.60) kJ = 3353 J (6)
•
In this example the disagreement between the
mechanical work and the heat energy is found to
be no more than about 1%. Due to unavoidable
tolerances relating to the composition of materi-
als (aluminum is very soft and almost impossible
to work mechanically, so that it is always alloyed),
the specific heat capacity can fluctuate quite no-
ticeably. The specific heat capacity is most easily
calculated by heating it electrically using the
equivalence between heat and electrical energy.
4.2 Conversion of electrical energy into heat
4.2.1 Experiment procedure
•
After cooling the friction cylinder it should be
screwed into the base (the same experimental
conditions as for the friction experiment) and the
temperature sensor inserted. After a few minutes
that should be ignored for the sake of homog-
enous distribution of temperature, the resistance
of the temperature sensor is R
1
= 8.00 k
Ω
(corre-
sponding to T
1
= 1460 °C by Eq. 1).
•
Now the power supply that has been configured
in advance (see Section 3) should be connected to
the heating element and a stopwatch started.
Voltage and current (as displayed by the power
supply) should be noted: U = 11.0 V, I = 0.510 A
•
After t = 600 s the experiment is halted and the
resistance of the sensor is read off: R
2
= 3.98 k
Ω
(T
2
= 30.32 °C).
4.2.2 Experiment evaluation
•
The electrical energy E is the product of power P
and time t. The power is the product of voltage
and current. Therefore:
E U T = = 11,0 x 0,512 x 600 = 3379Ws I (7)
•
In this experiment, the heat added is
∆
Q = c
A
m
A
(T
2
– T
1
) = 0.86 x 0.249
x (30.32 - 14.60) kJ = 3366J (8)
•
The agreement between E and
∆
Q is very good in
this instance as well.