The two ends of a metal rod are maintained at temperatures 100$^{\circ}$C and 110$^{\circ}$C. The rate of heat flow in the rod is found to be 4.0 J/s. If the end are maintained at temperatures 200$^{\circ}$C and 210$^{\circ}$C, the rate of heat flow will be

Question

Sphere of color black, red, white, and yellow are heated to the same temperature. The decreasing order of cooling is:

Answer 1

To determine the rate of heat flow when the ends of the metal rod are at different temperatures, we start by considering the basic principles of thermal conduction. According to Fourier's Law of Heat Conduction, the rate of heat flow ($Q/t$) through a material is given by the formula:

\frac{Q}{t} = \frac{k \cdot A \cdot \Delta T}{L}

k is the thermal conductivity of the material.
A is the cross-sectional area of the rod.
\Delta T is the temperature difference between the two ends.
L is the length of the rod.

Initially, the ends of the rod are at 100°C and 110°C, respectively, giving a temperature difference ($\Delta T$) of 10°C. The rate of heat flow ($Q/t$) is measured as 4.0 J/s under these conditions.

Now, if the temperatures are changed to 200°C and 210°C, the temperature difference ($\Delta T$) remains the same at 10°C. Since Fourier's Law indicates that the rate of heat flow depends linearly on the temperature difference, and other factors such as thermal conductivity, cross-sectional area, and length remain unchanged, the rate of heat flow will be the same.

Thus, under the new set of temperatures (200°C and 210°C), the rate of heat flow remains 4.0 J/s.

This solution can be confirmed using the proportionality provided by Fourier’s Law:

\frac{Q_1}{Q_2} = \frac{\Delta T_1}{\Delta T_2} ,

where $Q_1$ and $Q_2$ are the rates of heat flow, and $\Delta T_1$ and $\Delta T_2$ are the initial and final temperature differences:

\Delta T_1 = 10 \,^{\circ}C
\Delta T_2 = 10 \,^{\circ}C

The proportionality confirms that:

\frac{4.0}{Q_2} = \frac{10}{10} ,

implying $Q_2 = 4.0$ J/s. Therefore, the option 4.0 J/s is the correct answer.

Answer 2