The following table shows data collected to measure the thermal conductivity of steel specimen of 30 mm diameter: Specimen Length t lw ( C) Heater Heater H t 2w Qw t1 t2 K (mm) (V) (A) (W) (kg/s) (C) (W) ( C) (C) (W/m. K) Steel 60 100 0.65 20 23 65 49 i. Use the data provided above to complete the missing information in the table. (use V=0.5 L, t=123s) ii. Plot the variation of temperature with distance using the data presented in the table. Explain the relationship you obtain. Additional information: = 998 kg/m³ p water

Elements Of Electromagnetics
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The following table shows data collected to measure the thermal conductivity of steel specimen of 30 mm diameter:
Specimen
Length
(mm)
t1
(C)
Heater
Heater
Qw
t lw
( C)
t 2w
( C)
H
K
(V)
(A)
(W/m.
( C)
K)
(W)
(kg/s)
(W)
Steel
60
100
0.65
20
23
65
49
i. Use the data provided above to complete the missing information in the table. (use V=0.5 L , t=123s)
ii. Plot the variation of temperature with distance using the data presented in the table. Explain the relationship
you obtain.
Additional information:
= 998 kg/m3
p water
Cp = 4180 J/Kg.K
Transcribed Image Text:The following table shows data collected to measure the thermal conductivity of steel specimen of 30 mm diameter: Specimen Length (mm) t1 (C) Heater Heater Qw t lw ( C) t 2w ( C) H K (V) (A) (W/m. ( C) K) (W) (kg/s) (W) Steel 60 100 0.65 20 23 65 49 i. Use the data provided above to complete the missing information in the table. (use V=0.5 L , t=123s) ii. Plot the variation of temperature with distance using the data presented in the table. Explain the relationship you obtain. Additional information: = 998 kg/m3 p water Cp = 4180 J/Kg.K
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