Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# A Body Cools Down from 50°C to 45°C in 5 Mintues and to 40°C in Another 8 Minutes. Find the - Physics

Sum

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 T0°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 - t0]........... (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 − T0)°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 - T0] ........ (2)

Dividing (1) by (2),

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

⇒ 42.5 - T_0 = 29.68 - 0.625T

⇒ T0 = 34° C

Concept: Qualitative Ideas of Blackbody Radiation
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 6 Heat Transfer
Q 51 | Page 102