Question

A metal ball 0.1m in radius is heated from 273to348Kelvin.calculate the increase in surface area of ball (given beta=3.4×10^-5/kelvin

Answer 1

The increase in surface area of ball is 3.2 $cm^{2}$.

Explanation:

1. Radius of ball $(R_{o})$ = 0.1 m

  So initial surface area of ball $A_{o}=4\times \pi \times R_{o}^{2}$

2. Increase in temperature $(\Delta T)=T-T_{o}$     ...1)

  where

  T = final temperature = 348 K

 $T_{o}$ = initial temperature = 273 K

3. Final radius of ball = R

    Final surface area of ball = A

4. Area thermal expansion coefficient $(\beta )= 3.4\times 10^{-5} K^{-1}$

5.  Now from relation

   $A= A_{o}(1+\beta \times \Delta T)$      

   Which can also be written as

   $A-A_{o}= A_{o}\times \beta \times \Delta T$   ...2)  

6.  Increase in surface area = $A-A_{o}= A_{o}\times \beta \times \Delta T$

  So

  $A-A_{o}= 4\times \frac{22}{7}\times 0.1^{2}\times 3.4\times 10^{-5} \times (348-273)$    

 on solving

 $A-A_{o}=3.2\times 10^{-4} m^{2}= 3.2 cm^{2}$