A rod with negligible heat capacity has a length of 20 cm , a cross-sectional area of 1.0 cm^2 , and a thermal conductivity of 200 W m^ -1 ^ ^ ^ ^ -1 . One end's temperature is maintained at 0^ ^ ^ C , while the

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Quantrex Academy · Accepted Answer

Correct answer: A. 1800 J

The rate of heat transfer through the rod is given by Fourier's law of heat conduction:

\dfrac{dQ}{dt} = \dfrac{KA}{L} \Delta T

Since the temperature of one end is maintained at

0^{}^{{}^{\circ}}\text{C}

and the other end varies linearly from

0^{}^{{}^{\circ}}\text{C}

to

60^{}^{{}^{\circ}}\text{C}

over

10 \text{ minutes}

(

600 \text{ s}

), the temperature difference

\Delta T

as a function of time

t

is

\Delta T(t) = \dfrac{60}{600} t = 0.1 t

.

The total heat transmitted

Q

over the time interval

t = 0

to

t = 600 \text{ s}

is:

Q = \int_{0}^{600} \dfrac{KA}{L} \Delta T(t) dt

Substituting the given values

K = 200 \text{ W m}^{-1} ^{}^{{}^{\circ}}\textC^{-1}

,

A = 1.0 \times 10^{-4} \text{ m}^2

, and

L = 0.2 \text{ m}

:

\dfrac{KA}{L} = \dfrac{200 \times 10^{-4}}{0.2} = 0.1 \text{ W }^{}^{{}^{\circ}}\textC^{-1}

Therefore, the integral becomes:

Q = \int_{0}^{600} 0.1 \times 0.1 t dt = 0.01 \int_{0}^{600} t dt

Q = 0.01 \left[ \dfrac{t^2}{2} \right]_{0}^{600}

Q = 0.01 \times \dfrac{360000}{2} = 1800 \text{ J}

A rod with negligible heat capacity has a length of 20 cm , a...

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