A body cools down from 50°C to 45°C in 5 mintues and to 40°C in another 8 minutes. Find the temperature of the surrounding.

#### Solution

Let the temperature of the surroundings be T_{0}°C.

Case 1:

Initial temperature of the body = 50°C

Final temperature of the body = 45°C

Average temperature = 47.5 °C

Difference in the temperatures of the body and its surrounding = (47.5 − T)°C

Rate of fall of temperature = `(Deltat)/t = (5)/5 = 1°`C/min

By Newton's law of cooling,

`(dt)/dt = -K [T_{avg} - T_0 ]`

1 = -K [47.5 - t_{0}]........... (i)

Case 2:

Initial temperature of the body = 45°C

Final temperature of the body = 40°C

Average temperature = 42.5 °C

Difference in the temperatures of the body and its surrounding = (42.5 − T_{0})°C

Rate of fall of temperature = `(DeltaT)/t = 5/8 = 5/8`°C/min

From Newton^{,}s law of cooling,

`(dT)/(dt) = -K[T_{avg} - T_0 ]`

0.625 = -K [42.5 - T_{0}] ........ (2)

Dividing (1) by (2),

`1/0.625 = (47.5 - T_0)/(42.5 - T_0)`

⇒ 42.5 - T_0 = 29.68 - 0.625T

⇒ T_{0} = 34° C