Statement:
According to Newton’s law of cooling, the rate of loss if heat energy Q is directly proportional to the temperature difference between the temperature of the body and the surrounding.
Explanation:
- Let T be the temperature of the body which is greater than the temperature of the surrounding To.
- Then according to Newton’s law of cooling,
Rate if loss of heat α (T – To)
Hence, dQ/dt = K (T – To)
- Where K is the constant of proportionality.
- If m is the mass of the body and S be the specific heat of the body then rate of loss of heat can be expressed as
- Rate of loss of heat = mass*specific heat* rate of loss of temperature
- Thus, dQ/dt = m*S*dT/dt
But we have dQ/dt = K(T – To)
- Thus, m*S*dT/dt = K (T – To)
Thus, dT/dt = K/mS * (T – To)
- Thus, we can write if K, m and S are supposed to be constant then
dT/dt α (T – To)
Alternate statement of Newton’s law of cooling:
- The rate of fall of temperature is directly proportional to the temperature difference between the body and the surrounding, provided that the temperature difference is small.
Limitations of Newton’s law of cooling:
- Newton’s law of cooling can be applicable only when there is temperature difference between the body and the surrounding and that temperature difference is small.
- Newton’s law of cooling is not the law of radiation because here energy also lost by the process of conduction and convection.
Example:
If an object cools from 70°C to 60°C in 5 minutes and again cools to 50°C in next 6 minutes then what will be the temperature of the surrounding?
Solution:
Given that,
T1 = 70+60/2 = 130/2= 65°C
T2 = 60+50/2 = 110/2 = 55°C
Then
dT1/dt = 70-60/5 = 10/5 = 2°C/min
And dT2/dt = 60-50/6= 10/6= 5/3= 1.66°C/min
Then by Newton’s law of cooling,
(dT1/dt)/(dT2/dt) = (T1 – To)/(T2 – To)
Thus, 2/1.66= (65 – To)/(55 – To)
1.2*(55 – To) = 65 – To
66 – 1.2 To = 65 – To
66 – 65 = (1.2 – 1)*To
1 = 0.2*To
Thus, To = 1/0.2 = 5°C
Thus, the temperature of surrounding or the room temperature will be found as 5°C.