One end of a 0.25-meter copper rod with a cross-sectional area of 1.2×10^4 m² is

Question

One end of a 0.25-meter copper rod with a cross-sectional area of 1.2×10^4 m² is driven into the center of a sphere of ice at 0 °C (radius = 0.15 meters). The portion of the rod that is embedded in the ice is also at 0 °C. The rod is horizontal and its other end is fastened to a wall in a room. The rod and the room are kept at a constant temperature of 24 °C. The emissivity of the ice is 0.90. What is the ratio of the heat per second gained by the sphere through conduction to the net heat per second gained by the ice due to radiation? Neglect any heat gained thorugh the sides of the rod.

Explanation / Answer

We calculate the heat gained by ice sphere conduction

P= Q/t = k A dT/dx

data

A =1.2 10-4 m²

k= 397 W /m C (copper)

dT/dx = (Th – Tc) / L

indicate that as part of the rod within the ice is 0C, this does not contribute to the temperature gradient

dT/dx = (24 – 0) /(0.25-0.15)

dT/dx= 240 C/m

P= 397 1.2 10-4240

Pc= 11.43 W

Calor ganado por radiación

P = A e (Troom4 –Tice4)

Data

Troom = 24 C = 24+273 =297 K

T ice = 0 C = 273 K

Pr= 5.669 10-81.2 10-40.9 (2974 –2734)

Pr = 6.12 10-12(7.78 109- 5.55 109)

Pr = 13.65 10-3W

relationship between the two magnitudes

Pc/Pr = 11.43/13.65 10-3

Pc/Pr = 0.837 103

Pc/Pr = 837

This means that heat gain by conduction at 837 times the radiation gain

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