calculations: de Slope= dt 46-56 -10 -0.5 30–10 20 de From the graph, slope = Rate of cooling = -0.5 °C/min dt = -0.5 × 60 = -30 °C/sec The Thermal Conductivity constant may be calculated as de m xc x xd 0.803kg × 380 j/kg.k x 30 °C/sec x 1.70×10-3m k = at = 90.4809 W /m.K A(T1-T2) 9.5552×10-³m² x( 346–328) Where c= Specific heat capacity of brass = 380 j/kg.k K= 90.4809 W/m .K

College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Experiment :LEE’S DISC

I want the Analysis for this experiment 

calculations:
de
46-56
-10
Slope=
dt
-0.5
30-10
20
de
From the graph,
dt
slope
= Rate of cooling
-0.5 °C/min
— — 0.5 х 60 3D —30 °C/sec
The Thermal Conductivity constant may be calculated as
de
т хс х ха 0.803kg x 380 j/kg.k x 30 °C/sec X 1.70x10-3 т
k =-
A(T1-T2)
= 90.4809 W /m.K
9.5552x10-3m² ×( 346–328)
Where c= Specific heat capacity of brass = 380 j/kg.k
K= 90.4809 W /m.K
Transcribed Image Text:calculations: de 46-56 -10 Slope= dt -0.5 30-10 20 de From the graph, dt slope = Rate of cooling -0.5 °C/min — — 0.5 х 60 3D —30 °C/sec The Thermal Conductivity constant may be calculated as de т хс х ха 0.803kg x 380 j/kg.k x 30 °C/sec X 1.70x10-3 т k =- A(T1-T2) = 90.4809 W /m.K 9.5552x10-3m² ×( 346–328) Where c= Specific heat capacity of brass = 380 j/kg.k K= 90.4809 W /m.K
Temperature vs Time graph
70
60
50
40
30
20
10
10
15
20
25
30
35
Time (min)
Temperature (°C)
Transcribed Image Text:Temperature vs Time graph 70 60 50 40 30 20 10 10 15 20 25 30 35 Time (min) Temperature (°C)
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