Question

A pair of eyeglass frames are made of an epoxy plastic (coefficient of linear expansion = 122 ✕ 10-6 °C-1). At room temperature (20.0°C) the frames have circular lens holes 2.00 cm in radius. To what temperature must the frames be heated if lenses 2.01 cm in radius are to be inserted into them?

Answer 1

Answer:

The new temperature is  $T_1 = 60.78^oC$

Explanation:

From the question we are told that

   The coefficient of linear expansion is  $\sigma = 122 *10^{-6} \ ^oC^{-1}$

    The temperature is  $T = 20.0 ^oC$

    The radius of the frames is $r = 2.00 \ cm = 0.02 \ m$

    The new radius is  $r_2 = 2.01 \ cm = 0.021 \ m$

The change in radius is mathematically represented as

       $\Delta r = r_1 -r$

substituting values

       $\Delta r = 2.01 - 2.00$

       $\Delta r = 0.01 \ m$

The increase in radius can also be mathematically represented as

     $\Delta r = r * \sigma (T_1 -T)$

Where $T_1$ is the the new temperature , making it the subject we have

      $T_1 = \frac{\Delta r}{r * \sigma } + T$

substituting value  

     $T_1 = \frac{0.01}{2.01 *122*10^{-6} } + 20$

    $T_1 = 60.78^oC$