State and explain newton's law of cooling
Answers
Answered by
19
curve.
Theory
This Law of Cooling is named after the famous English Physicist Sir Isaac Newton, who conducted the first experiments on the nature of cooling.
Statement of the Law :
According to Newton’s Law of Cooling, the rate of cooling of a body is directly proportional to the difference in temperatures of the body (T) and the surrounding (T0), provided difference in temperature should not exceed by 300C.
From the above statement,

For a body of mass m, specific heat s, and temperature T kept in surrounding of temperature T0;

Now, the rate of cooling,

Hence ,

Since the mass and the specific heat of the body are taken as constants, the rate of change of temperature with time can be written as,

The above equation explains that, as the time increases, the difference in temperatures of the body and surroundings decreases and hence, the rate of fall of temperature also decreases.
Theory
This Law of Cooling is named after the famous English Physicist Sir Isaac Newton, who conducted the first experiments on the nature of cooling.
Statement of the Law :
According to Newton’s Law of Cooling, the rate of cooling of a body is directly proportional to the difference in temperatures of the body (T) and the surrounding (T0), provided difference in temperature should not exceed by 300C.
From the above statement,

For a body of mass m, specific heat s, and temperature T kept in surrounding of temperature T0;

Now, the rate of cooling,

Hence ,

Since the mass and the specific heat of the body are taken as constants, the rate of change of temperature with time can be written as,

The above equation explains that, as the time increases, the difference in temperatures of the body and surroundings decreases and hence, the rate of fall of temperature also decreases.
Answered by
36
Newton's law of cooling: The rate of loss of heat, -dQ/dt of the body is directly proportional to the difference of temperature /\T=(T2-T1) of the body and the surroundings.
Mathematically:
-dQ/dt= k(T2-T1)--------(1)
Rate of loss of heat is given by
dQ/dt =ms dT2/dt-------(2)
From eq (1) and (2) we have
-msdT2/dt = k (T2-T1)
dT2/T2-T1=-k/ms (dt) =-K st
Where K=k/ms
On integrating,
loge(T2-T1)=-Kt+c
(T2-T1)=e^-Kt+c
Hope it is useful........
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