Question

An iron rod of mass \(2\,\text{kg}\) and length \(1.5\,\text{m}\) is supplied \(18.4\,\text{kJ}\) of heat. If the specific heat capacity of iron is \(460\,\text{J kg}^{-1}\text{K}^{-1}\) and coefficient of linear expansion is \(1.2 \times 10^{-5}\^\circ\text{C}^{-1}\), then the increase in length of the rod is: 

• A. \(0.24\,mm\)
• B. \(0.36\,mm\)
• C. \(0.12\,mm\)
• D. \(0.18\,mm\)

Hint:

Always find the temperature rise first using \[ Q=ms\Delta T, \] then substitute into \[ \Delta L=\alpha L\Delta T. \]

Answer 1

The Correct Option is B

Concept: Heat supplied raises the temperature: \[ Q=ms\Delta T. \] Linear expansion is \[ \Delta L=\alpha L\Delta T. \]

Step 1: Find temperature rise.
\[ Q=18.4kJ=18400J. \] \[ 18400 = 2\times460\times\Delta T. \] \[ 18400 = 920\Delta T. \] \[ \Delta T=20^\circ C. \]

Step 2: Apply linear expansion formula.
\[ \Delta L = \alpha L\Delta T. \] \[ = (1.2\times10^{-5})(1.5)(20). \] \[ = 3.6\times10^{-4}m. \]

Step 3: Convert into millimetres.
\[ \Delta L = 0.36mm. \] Hence \[ \boxed{0.36\,mm}. \]