Question: In a room where the temperature is 300C, a body cools from 610C to 590</s...

Question

In a room where the temperature is 30⁰C, a body cools from 61⁰C to 59⁰C is 4 minutes. The time taken by the body to cool from 51⁰C to 49⁰C will be:

Answer 1

Rate of cooling $\alpha$ difference in temperature

$\frac{dT}{dt}\alpha\Delta\theta$ ⇒ $\frac{dT}{dt} = K\Delta\theta$

in first case

$dT = 61 - 59 = 2$

$\Delta\theta = 60 - 30 = 30$

dt = 4 minute

$\therefore$ $K = \frac{dT}{\Delta\theta dt} = \frac{2}{30x4} = \frac{1}{60}$

For second case

$dT = 2$

∆θ = 50 − 30 = 20

$\therefore$ $dt = \frac{dT}{K\Delta\theta} = \frac{2}{\frac{1}{60}x20} = 6min.$ Hence (2) is correct

Question: In a room where the temperature is 30<sup>0</sup>C, a body cools from 61<sup>0</sup>C to 59<sup>0</s...