A piece of steel has a length of 30 cm at 15°C. At 90°C its length increases by 0.027 cm. Coefficient of linear expansion of steel piece is:
6 × 10-6°C-1
Calculating the Coefficient of Linear Expansion of Steel
This problem involves calculating the coefficient of linear expansion for a piece of steel given its initial length, the change in temperature, and the resulting increase in length. The principle of linear thermal expansion is used here.
The formula that describes linear thermal expansion is:
\[ \Delta L = L_0 \alpha \Delta T \]Where:
- \(\Delta L\) is the change in length
- \(L_0\) is the original length
- \(\alpha\) is the coefficient of linear expansion (which we need to find)
- \(\Delta T\) is the change in temperature
From the problem statement, we are given the following values:
- Initial length, \(L_0 = 30 \text{ cm}\)
- Initial temperature, \(T_1 = 15^\circ\text{C}\)
- Final temperature, \(T_2 = 90^\circ\text{C}\)
- Increase in length (change in length), \(\Delta L = 0.027 \text{ cm}\)
First, let's calculate the change in temperature, \(\Delta T\).
\[ \Delta T = T_2 - T_1 = 90^\circ\text{C} - 15^\circ\text{C} = 75^\circ\text{C} \]Now, we need to rearrange the linear expansion formula to solve for the coefficient of linear expansion, \(\alpha\):
\[ \alpha = \frac{\Delta L}{L_0 \Delta T} \]Next, we substitute the given values into this rearranged formula:
\[ \alpha = \frac{0.027 \text{ cm}}{30 \text{ cm} \times 75^\circ\text{C}} \]Let's perform the multiplication in the denominator:
\[ 30 \times 75 = 2250 \]So, the formula becomes:
\[ \alpha = \frac{0.027}{2250} \text{ }^\circ\text{C}^{-1} \]To simplify the calculation, we can write 0.027 as \(27 \times 10^{-3}\):
\[ \alpha = \frac{27 \times 10^{-3}}{2250} \text{ }^\circ\text{C}^{-1} \]Now, divide 27 by 2250:
\[ \frac{27}{2250} = \frac{27 \times 10}{2250 \times 10} = \frac{270}{22500} \]We can simplify this fraction. Both are divisible by 9:
\[ \frac{270 \div 9}{22500 \div 9} = \frac{30}{2500} \]Now, divide by 10:
\[ \frac{30 \div 10}{2500 \div 10} = \frac{3}{250} \]To convert this to a decimal, divide 3 by 250:
\[ \frac{3}{250} = \frac{3 \times 4}{250 \times 4} = \frac{12}{1000} = 0.012 \]So, returning to the calculation for \(\alpha\):
\[ \alpha = 0.012 \times 10^{-3} \text{ }^\circ\text{C}^{-1} \]Expressing this in standard scientific notation:
\[ \alpha = 1.2 \times 10^{-2} \times 10^{-3} \text{ }^\circ\text{C}^{-1} \] \[ \alpha = 1.2 \times 10^{-5} \text{ }^\circ\text{C}^{-1} \]Let's re-calculate $\frac{0.027}{2250}$ directly using powers of 10:
\[ \alpha = \frac{2.7 \times 10^{-2}}{2.25 \times 10^3} \text{ }^\circ\text{C}^{-1} \] \[ \alpha = \frac{2.7}{2.25} \times 10^{-2} \times 10^{-3} \text{ }^\circ\text{C}^{-1} \] \[ \alpha = \frac{2.7}{2.25} \times 10^{-5} \text{ }^\circ\text{C}^{-1} \] \[ \frac{2.7}{2.25} = \frac{270}{225} = \frac{54 \times 5}{45 \times 5} = \frac{54}{45} = \frac{6 \times 9}{5 \times 9} = \frac{6}{5} = 1.2 \] \[ \alpha = 1.2 \times 10^{-5} \text{ }^\circ\text{C}^{-1} \]This result can also be written as \(12 \times 10^{-6} \text{ }^\circ\text{C}^{-1}\).
Therefore, the coefficient of linear expansion of the steel piece, based on the given data, is \(12 \times 10^{-6} \text{ }^\circ\text{C}^{-1}\).
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 :
How much should the temperature of a brass rod be increased so as to increase its length by 1%?
Given: for brass α = 0.00002/°CA 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?
