PASCO TD-8552 ELECTRICAL EQUIVALENT OF HEAT - Manual (Page 10)

Electrical Equivalent of Heat 012-02833D
6
Calculations
In order to determine the efficiency of the lamp, it is necessary to determine both the total
electrical energy that flowed into the lamp (E) and the total heat absorbed by the water (H).
E, the electrical energy delivered to the lamp:
E = Electrical Energy into the Lamp = V . I . t = __________________________
t = t
f
- t
i
= the time during which power was applied to the lamp = ________
H, the heat transferred to the water (and calorimeter):
H = (M
w
+M
e
)(1 cal/gm C)(T
f
-T
i
) = __________________________________
M
w
= M
jw
- M
j
= Mass of water heated = ____________________________
H
j
= H J
e
= ____________________________________________________
M
e
= 23 grams. Some of the heat produced by the lamp is absorbed by the EEH Jar. For
accurate results, therefore, the heat capacity of the jar must be taken into acount (The heat
capacity of the EEH Jar is equivalent to that of approximately 23 grams of water.)
Efficiency:
E-H
j
= _________________________________________________________
E
Questions
➀ What effect are the following factors likely to have on the accuracy of your determination of
the efficiency of the lamp? Can you estimate the magnitude of the effects?
a. Water is not completely transparent to visible light.
b. Not all the infrared radiation is absorbed by the water.
c. The styrofoam Calorimeter was not used, so there is some transfer of thermal energy
between the EEH Jar and the room atmosphere.
➁ Is an incandescent lamp more efficient as a light bulb or as a heater?