Question: In a room where temperature is 30oC a body cools from 61oC to 59oC ...

Question

In a room where temperature is 30^oC a body cools from 61^oC to 59^oC is 4 minutes. The time taken by the body to cool from 51^oC to 49^oC will be:

Answer 1

Solution

Rate of cooling $\alpha$ difference in temperature

−

dT = − K∆θ.dt

In First Case

dT = 61−59=2

∆θ = 60 − 30 = 30

dt = 4 minutes

$\therefore$ $\mathrm { K } = - \frac { \mathrm { dT } } { \Delta \theta \mathrm { dt } } = - \frac { 2 } { 30 \mathrm { x } 4 } = - \frac { 1 } { 60 }$

For second case dT = 2

$\Delta \theta = 50 - 30 = 20$

$\therefore$

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

Solution