Question

A cup of tea cools from 65° to 62 degrees in one minute if at room temperature 22.5° water how much time will it take from to cool it from 46 degree 40 degree

Answer 1

According to Newton's law of cooling,

$\frac{dT}{dt}=k(T-T_s)$

here, dT is change in temperature ,

dt is change in time.

T is taken to be average since changes in temperature is less than s, $T_s$ = 22.5°C

case 1 : dT = (65 - 62) = 3°C

dt = 1 min

and T = (65 + 62)/2 = (127)/2 = 63.5°C

now, applying Newton's law of cooling,

3/1 = k(63.5 - 22.5)

k = 3/41 ....... (1)

case 2 : dT = (46 - 40) = 6°C

dt = ?,

and T = (46 + 40)/2 = (86)/2 = 43°C

now again applying Newton's law of cooling,

6/dt = k(43 - 22.5)

from equation (1),

or, 6/dt = (3/41) × 41/2

or, 6/dt = 3/2

dt = 4 min

hence, time taken to cool from 46° to 40°C is 4 min