A welding rod with κ = 30 (Btu/hr)/(ft ⋅ °F) is 20 cm long and has a diameter of 4 mm. The two ends of the rod are held at 500 °C and 50 °C. (a) In the units of Btu and J, how much heat flows along the rod each second? (b) What is the temperature of the welding rod at its midpoint?

In Btu:

Q=0.001390 Btu.

In Joule:

Q=1.467 J

Part B:

Temperature at midpoint=274.866 C

Explanation:

Thermal Conductivity=k=30  (Btu/hr)/(ft ⋅ °F)=

Thermal Conductivity is SI units:

Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft

Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft

T_1=500 C=932 F

T_2=50 C= 122 F

Part A:

In Joules (J)

Heat Q is:

In Btu:

Heat Q is:

PArt B:

At midpoint Length=L/2=0.1 m

On rearranging:

1

explanation:

answer:

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explanation: