How much should the temperature of a brass rod be increased so as to increase its length by 1%?
500°C
Calculating Temperature Increase for Brass Rod Length Expansion
This problem involves the concept of linear thermal expansion, which describes how the dimensions of a material change with temperature.
When a material is heated, its particles vibrate more vigorously and move farther apart on average, leading to an increase in size. For a one-dimensional object like a rod, this expansion is primarily along its length, known as linear expansion.
The formula for linear thermal expansion is given by:
\[ \Delta L = L_0 \alpha \Delta T \]Where:
- \( \Delta L \) is the change in length.
- \( L_0 \) is the original length.
- \( \alpha \) is the coefficient of linear expansion for the material.
- \( \Delta T \) is the change in temperature.
The question asks for the temperature increase \( \Delta T \) required to increase the length of a brass rod by 1%. This means the change in length \( \Delta L \) is 1% of the original length \( L_0 \).
Given:
- Percentage increase in length = 1%
- This translates to \( \frac{\Delta L}{L_0} = 1\% = \frac{1}{100} = 0.01 \)
- Coefficient of linear expansion for brass, \( \alpha = 0.00002 /^\circ\text{C} = 2 \times 10^{-5} /^\circ\text{C} \)
We need to find \( \Delta T \). We can rearrange the linear expansion formula to solve for \( \Delta T \):
\[ \Delta T = \frac{\Delta L}{L_0 \alpha} \]Now, substitute the given values into the rearranged formula:
\[ \Delta T = \frac{\left(\frac{\Delta L}{L_0}\right)}{\alpha} \] \[ \Delta T = \frac{0.01}{2 \times 10^{-5} /^\circ\text{C}} \] \[ \Delta T = \frac{10^{-2}}{2 \times 10^{-5}} \ ^\circ\text{C} \] \[ \Delta T = \frac{1}{2} \times \frac{10^{-2}}{10^{-5}} \ ^\circ\text{C} \] \[ \Delta T = 0.5 \times 10^{-2 - (-5)} \ ^\circ\text{C} \] \[ \Delta T = 0.5 \times 10^{3} \ ^\circ\text{C} \] \[ \Delta T = 0.5 \times 1000 \ ^\circ\text{C} \] \[ \Delta T = 500 \ ^\circ\text{C} \]The temperature increase required is 500 degrees Celsius.
Note that a change in temperature of 1 degree Celsius is equivalent to a change in temperature of 1 Kelvin. Therefore, \( \Delta T = 500 \ ^\circ\text{C} \) is the same magnitude of temperature change as \( \Delta T = 500 \) K. However, the coefficient of expansion is given in $^\circ\text{C}^{-1}$, and the options include values in $^\circ\text{C}$.
Comparing our calculated value with the given options:
| Option | Temperature Increase |
|---|---|
| 1 | 1000 K |
| 2 | 1000°C |
| 3 | 500 K |
| 4 | 500°C |
Our calculated value of \( 500 \ ^\circ\text{C} \) matches Option 4.
Important Questions from Thermal expansion
Thermal expansion of solids are:
A copper rod and a steel rod are to have lengths LC and LS, such that the difference between their lengths is the same at all ambient temperatures. If the coefficients of linear expansion of copper and steels are αC and αS respectively. The lengths are related to the coefficient of linear expansion as :
A wooden wagon wheel has an outside diameter of 3750 mm. The iron tire for this wheel is deliberately made smaller so that it can be shrunk in place to be a tight fit. If the tire's inside diameter is 3737 mm at 20°C, the temperature to which it must be heated to fit over the wheel? The coefficient of linear expansion of the steel is 1.2 × 10-5/°C.
A cylinder of cross-sectional radius 1 cm and height 4 cm is heated from 0°C to 100°C. If the coefficient of linear expansion α = 4 × 10-4/°C, what will be the increase in the volume of the cylinder?
Greater the value of _______ of a material, the more rapidly it will conduct heat.
