Question

The temperature of a body falls from 50^(@)C to 40^(@)C in 10 minutes. If the temperature of the surroundings is 20^(@)C Then temperature of the body after another 10 minutes will be

Answer 1

Given:

Initial temperature $\theta _{i} =$ 50°C

Final temperature $\theta _{s} =$ 20°C

The time take to temperature drop $dt = 10$ min

To Find:

Temperature of the body after another 10 minutes will be.

According to the newton law of cooling,

    $\frac{d\theta }{dt} = -K (\theta _{i} - \theta _{s} )$

Where $K =$ cooling constant

  $\frac{-10}{10} = -K (50-20)$

    $K = \frac{1}{30}$

For finding temperature of the body after 10 minutes,

   $\frac{T-40}{10} = -\frac{1}{30} (40-20)$

 $T -40 = -6.66$

 $T = 40 -6.66$

 $T =$ 33.3°C

Therefore, the temperature of the body after another 10 minutes will be 33.3°C