Q1
ΔQ/Δt =- κ•A•ΔT/L = >
ΔQ =- κ•A•ΔTt/L=
=- 50.4•0.1•(-20) •30•60/0.02 =
= 9072000 ≈9100000 J
Q2
q= ΔQ/Δt =- κ•A•ΔT/L
κ = - q•L/ A•ΔT =
= - 4•q•L/ π•d²•ΔT=
=- 4• 164.9•0.3/3.14•(0.025)² •(20-500) =
=209.9 W/m•℃
(Q1)determine the quantity of heatconducted in 30min through an iron plate2.0cm thick and 0.10m2 in area if the temperature of the two sides are 0c and 20c.the coefficent of thermal conductivity is 50.4j/smc (a)91000000j(b)23000000j(c)41000000j(d)33000000j (Q2) one end of a 30-cm long aluminum rod is exposed to a temperature of 500c while the other is maintained at 20c.the rod has the diameter of 2.5cm. if heat is conducted through the rod at the rate of 164.9j/s. calculate the thermal conductivity of Aluminum(a)311.6w/mc(b)209.9w/mc(c)506.3w/mc(d)457.1w/mc
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