A body cools down from 50°C to 45°C in 5 minutes and then from 45°C to 40°C in another 8 minutes. The temperature of the surroundings is approximately

Concept:

According to Newton’s law of cooling, the rate of loss of heat from a body is directly proportional to the difference in the temperature of the body, and its surroundings.

Rate of cooling ∝ ΔT

\(\frac{{dT}}{{dt}} = - k\left( {T_{avg} - {T_0}} \right)\)

• The graph drawn between the temperature of the body and time is known as the cooling curve.
• The plot for the rate of change of temperature is steeper in the beginning and as the temperature decreases with time, it becomes gradual; hence the rate of cooling is higher at the beginning and decreases as the temperature of the body falls.

Explanation:

Let the temperature of the surrounding be T₀°C

Case 1-

• The initial temperature of the body = 50°C
• The final temperature of the body = 45°C
• The average temperature of the body = 47.5°C
• Difference in temperature of the body and it is surrounding = (47.5 – T)°C
• Rate of fall of the temperature = △t/t = 10C/min

By newton’s law of cooling

\(\frac{{dT}}{{dt}} = - k\left( {T_{avg} - {T_0}} \right)\)

1 = - K [47.5 – t_o]     --- (1) 

Case 2 -

• Initial temperature of the body = 45°C
• Final temperature of the body = 40°C
• Average temperature of the body = 42.5°C
• Difference in temperature of the body and it is surrounding = (42.5 – T_o)°C
• Rate of fall of temperature = △ 5/8°C/min

From newton’s law of cooling

\(\frac{{dT}}{{dt}} = - k\left( {T_{avg} - {T_0}} \right)\)

0.625 = - K [ 42.5 – T_o]     --- (2)

Dividing equation (1) by (2)

1/0.625 = [47.5 – T_o] / [42.5 – T_o]

42.5 – T_o = 29.68 – 0.625 T

T_o = 34°C 

The correct answer to this question is 34°C.