A conductor of length L is placed along the x-axis, with one of its ends at x=0 and the other at x=L . If the rate of flow of heat energy through the conductor is constant and its thermal resistance per unit length is also constant, then which of the following graphs is/are correct?

Question

A conductor of length L is placed along the x-axis, with one of its ends at x=0 and the other at x=L . If the rate of flow of heat energy through the conductor is constant and its thermal resistance per unit length is also constant, then which of the following graphs is/are correct? (A) Option 1

(B)

(C)

(D)

Tardigrade Admin · Accepted Answer

We know that heat is flowing (d Q/d t)=-(K A (T - T0)/x) T=-((d Q/d t))(x/K A)+T0 Therefore the graph of the negative slope but positive intercept

Q. A conductor of length $L$ is placed along the x-axis, with one of its ends at $x = 0$ and the other at $x = L$ . If the rate of flow of heat energy through the conductor is constant and its thermal resistance per unit length is also constant, then which of the following graphs is/are correct?

We know that heat is flowing $\frac{d Q}{d t} = - \frac{K A ( T - T _{0} )}{x}$
$T = - (\frac{d Q}{d t}) \frac{x}{K A} + T_{0}$
Therefore the graph of the negative slope but positive intercept

Q. A conductor of length L is placed along the x-axis, with one of its ends at x=0 and the other at x=L . If the rate of flow of heat energy through the conductor is constant and its thermal resistance per unit length is also constant, then which of the following graphs is/are correct?

We know that heat is flowing dtdQ​=−xKA(T−T0​)​ T=−(dtdQ​)KAx​+T0​ Therefore the graph of the negative slope but positive intercept

Q. A conductor of length $L$ is placed along the x-axis, with one of its ends at $x = 0$ and the other at $x = L$ . If the rate of flow of heat energy through the conductor is constant and its thermal resistance per unit length is also constant, then which of the following graphs is/are correct?

We know that heat is flowing $\frac{d Q}{d t} = - \frac{K A ( T - T _{0} )}{x}$
$T = - (\frac{d Q}{d t}) \frac{x}{K A} + T_{0}$
Therefore the graph of the negative slope but positive intercept