Diameter of the disc = 110.3 mm Radius of the disc, r = .. .mm Area of disc = n r² = %3D .... 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 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 k = A(T1-T2) „equation 4 Where c= Specific heat capacity of brass=----- (Standard value)
Diameter of the disc = 110.3 mm Radius of the disc, r = .. .mm Area of disc = n r² = %3D .... 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 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 k = A(T1-T2) „equation 4 Where c= Specific heat capacity of brass=----- (Standard value)
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I want the calculation for this expiremnt
Note: the unit of thermal conductivity w/m.k
![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
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-
k =.
A(T-T2)
equation 4
Where c= Specific heat capacity of brass=--
(Standard value)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd467df0a-06aa-4201-9d7f-251867c68e3a%2Fb8ab2b29-ff07-4373-ba71-e111da05bc59%2F9qpn8ig_processed.png&w=3840&q=75)
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
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-
k =.
A(T-T2)
equation 4
Where c= Specific heat capacity of brass=--
(Standard value)
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