Question

In a certain experiment, a small radioactive source must move at selected, extremely slow speeds. This motion is accomplished by fastening the source to one end of an aluminum rod and heating the central section of the rod in a controlled way. If the effective heated section of the rod in Figure has length $\mathbf{d}\mathbf{=}\mathbf{2}\mathbf{.00}\mathbf{\text{\hspace{0.17em}cm}}$, at what constant rate must the temperature of the rod be changed if the source is to move at a constant speed of $\mathbf{100}\mathbf{\text{\hspace{0.17em}nm/s}}$?

Short answer

Rate of change of the temperature of the rod with respect to time is$0.2173°\text{\hspace{0.17em}C}/\text{s}$

Step-by-step solution

The given data 

1. Distance,$\text{L}=2.00\text{\hspace{0.17em}cmor}0.02\text{\hspace{0.17em}m}$
2. Speed of the source,$\text{v}=100\times {10}^{-9}\text{m}/\text{s}$
3. Thermal expansion coefficient for aluminum rod$\mathrm{\alpha }=23\times {10}^{-6}/°\text{\hspace{0.17em}C}$

Understanding the concept of thermal expansion

When an object's temperature changes, it expands and grows larger, a process known as thermal expansion. We know that the speed is the ratio of distance to time. In this experiment, the radioactive source is moving with a constant speed of 100nm/s. The source is radioactive, and the rod is heated as shown in figure. Therefore, thermal expansion takes place, and using the equation for linear expansion, we can find the rate of change of the temperature of the rod with respect to time.

Formula:

The linear expansion of a body,$\mathrm{\Delta }\text{L}=\text{L}\mathrm{\alpha }\mathrm{\Delta }\text{T}$ …(i)

where,$\mathrm{\alpha }$ is the coefficient of linear expansion of body.

Calculation of rate of change of time

Dividing the equation (i) by time increment$∆\mathrm{t}$ and equating it to the speed,$\text{v}=100\times {10}^{-9}\text{m}/\text{s}$we get

$\begin{array}{c}\frac{\mathrm{\Delta }\text{L}}{\mathrm{\Delta }\text{t}}=\frac{\text{L}\mathrm{\alpha \Delta }\text{T}}{\mathrm{\Delta }\text{t}}\\ \text{v}=\mathrm{\alpha }\text{L}\frac{\mathrm{\Delta }\text{T}}{\mathrm{\Delta }\text{t}}\text{(}\because \text{speedofthebody,v}=\frac{\mathrm{\Delta }\text{L}}{\mathrm{\Delta }\text{t}}\text{)}\\ \frac{\text{v}}{\mathrm{\alpha }\text{L}}=\frac{\mathrm{\Delta }\text{T}}{\mathrm{\Delta }\text{t}}\\ \frac{\mathrm{\Delta }\text{T}}{\mathrm{\Delta }\text{t}}=\frac{100\times {10}^{-9}\text{\hspace{0.17em}m}/\text{s}}{23\times {10}^{-6}/°\text{\hspace{0.17em}C}\times 0.02\text{\hspace{0.17em}m}}\text{\hspace{0.17em}(}\because \text{substitutingthegivenvalues)}\\ =0.2173°\text{\hspace{0.17em}C}/\text{s}\end{array}$

Hence, the constant rate of temperature is$0.2173°\text{\hspace{0.17em}C}/\text{s}$