Diameter of the disc = 110.3 mm Radius of the disc, r = ....mm ... Area of disc = Tn r' .. ........... Thickness of the disc, h = 10.5 mm Mass of the disc, m = 803 g Thickness of the experimental disc (say cardboard sheet), d = 1.70 mm Steady temperature of the steam chamber, T1 = 73 °C Steady temperature of the brass disc, T2 = 55 °C
Diameter of the disc = 110.3 mm Radius of the disc, r = ....mm ... Area of disc = Tn r' .. ........... Thickness of the disc, h = 10.5 mm Mass of the disc, m = 803 g Thickness of the experimental disc (say cardboard sheet), d = 1.70 mm Steady temperature of the steam chamber, T1 = 73 °C Steady temperature of the brass disc, T2 = 55 °C
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Experiment :LEE’S DISC
I want the graph and calculations for this experiment
Transcribed Image Text:Diameter of the disc = 110.3 mm
Radius of the disc, r = . ..mm
Area of disc = T r²
Thickness of the disc, h = 10.5 mm
%3D
............
%3D
Mass of the disc, m = 803 g
Thickness of the experimental disc (say cardboard sheet), d = 1.70 mm
Steady temperature of the steam chamber, T1 = 73 °C
Steady temperature of the brass disc, T2 = 55 °C
Time (min)
0 2
4
6 8
10 12 14 16 18 20 22 | 24 26 28
30
Temperature
59 | 58
57 56 56
56 55 54 53 52 51 50 49
48 47
46
(°C)
de
From the graph,
dt
slope = Rate of cooling
The Thermal Conductivity constant may be calculated as
de
m xc X
xd
k =
dt.
.equation 4
A(T-T2)
Where c= Specific heat capacity of brass=---
- (Standard value)
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