Question

Calculate the K using following observations . Specific of heat of material of slab, s = 910 Joule/kg ○C Mass of the Slab, M = 0.149 kg Radius of the disc , r = 4.2 cm Thickness of the disc, d = 1 mm Steady state temperature : T1 = 100 ○C T2= 84.25 ○C Temperature gradient, dT/dx = 0.142 ○C/sec​

Answer 1

Answer:

Heat Capacity =Specific heat×mass

=4200J/Kg

o

C×10

=42000J

=42kJ