Answer: a) 222k/m
b) 42.5W
c) if cold: 26.7C
If hot: 73.3C
Explanation:
If TH and TC are at the ends of the rod in the figure, then;
(A) 100 K / 45 cm = 2.22 K/cm or 222 K/m
(B) thermal conductivity constant for copper is 390 W/(m K).
Then the heat flow rate = Q/t
Q/t = [390 W/(m K)] (100 K) [ pi (0.0125 m)^2 ] / (0.45 m)
= 42.5 watts
(C) 12/45 = 4/15 = 0.2667
If the left end is the cold end, the answer will be 26.7 C
If the left end is the hot end, the answer will be 73.3 C
Following are solutions to the given points:
For point a)
Temperature gradient along the rods at the end of its steady-state.
[tex]\to \frac{\Delta T}{ d} = \frac{( 100^{\circ} \ C - 0^{\circ} \ C )}{(45 \ cm )}\\\\[/tex]
[tex]= 2.2222^{\circ} \ \frac{C}{cm}\\\\= 222.22 ^{\circ} \ \frac{ C}{m} \\\\= 222.22 \frac{K}{m}[/tex]
For point b)
The final steady-state heat current in the rod.
[tex]\to \frac{Q}{ t} = \frac{kcu A \Delta \ T}{ d}\\\\[/tex]
[tex]= \frac{\frac{ 385\ W}{(m.K)} \times 1.25 \times 10^{-4}\ m^2 \times 100\ K )}{(0.45 m )}\\\\= 10.69 \ W\\\\[/tex]
For point c)
A final steady-state temperature at 12.0 cm from the rod's left end.[tex]\to T_{12} = TH - (\frac{ \Delta T} {d} ) \times ( 0.12\ m )\\\\[/tex]
[tex]= 100^{\circ} \ C - ( 222^{\circ} \ \frac{C}{m} \times 0.12\ m )\\\\= 100^{\circ} \ C - 26.64^{\circ} \ C \\\\ = 73.36^{\circ} \ C \\\\\[/tex]
Find out more information about the temperature here:
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