Question

The ends of a cylindrical steel rod are maintained at two different temperatures. The rod conducts heat from one end to the other at a rate of 10 cal/s. At what rate would a steel rod twice as long and twice the diameter conduct heat between the same two temperatures?

Answer 1

Answer:20 cal/s

Explanation:

Given

Heat transfer rate is $\dot{Q}=10 cal/s$

Also heat rate is given by

$\dot{Q}=kA\frac{dT}{dx}$

where $k=thermal\ conductivity$

$A=area\ of\ cross-section$

$dT=change\ in\ temperature$

$dx=change\ in\ length$

$10=k\frac{\pi }{4}d^2\frac{dT}{L}----1$

For $d'=2d, Length L'=2L$

$\dot{Q}=k\frac{\pi }{4}(2d)^2\frac{dT}{2L}---2$

dividing 1 and 2 we get

$\frac{10}{\dot{Q}}=\frac{2d^2}{4d^2}$

$\dot{Q}=20 cal/s$                            

Answer 2

The power rate of the steel is mathematically given as

P=20cal/s

What rate would a steel rod twice as long and twice the diameter conduct heat between the same two temperatures?

Question Parameter(s):

The rod conducts heat from one end to the other at a rate of 10 cal/s

Generally, the equation for the Power   is mathematically given as

Power = (kA∆T)/l

Therefore

Power = [k × 4(πd^2)/4 × ∆T]/2l

Therefore

Power  = 2(kA∆T)/l

Hence, Initial Power

iP = (kA∆T)/l

iP= 10cal/s

Where

dT is the same

2[(kA∆T)/l] = 2 × 10

P=20cal/s

In conclusion, The power rate is

P=20cal/s

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