Write the newton law of cooling and explain
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Newton’s law of cooling describes the rate at which an exposed body changes temperature through radiation which is approximately proportional to the difference between the object’s temperature and its surroundings, provided the difference is small, i.e.,
The Newton’s law of cooling is given by,
dT/dt = k(Tt – Ts)
Where
Tt = temperature at time t and
Ts = temperature of the surrounding,
k = constant.
The Newton’s Law of Cooling Formula is expressed by
T(t) = Ts + (T0 – Ts) e-Kt
Where,
t = time,
T(t) = temperature of the given body at time t,
Ts = surrounding temperature,
To = initial temperature of the body,
k = constant.
Please note that greater the difference in temperature between the system and surrounding, more rapidly the body temperature changes.
Example 1
The oil is heated to 70oC. It cools to 50oC after 6 minutes. Calculate the time taken by the oil to cool from 50oC to 40oC given the Surrounding temperature Ts = 25oC.
Solution:
Given:
Temperature of oil after 10 min = 50oC,
Ts = 25oC,
To = 70oC,
t = 6 min
The Newton’s law of cooling formula is expressed as
T(t) = Ts + (To – Ts) e-kt
T(t)−Ts / To−Ts = e-kt
-kt ln = ln T(t)−Ts / To−Ts
-kt = ln 50−25 / 70−25
= ln 0.555
k = – (-0.555 / 6 )
= 0.092
If Tt = 45oC (average temperature as the temperature decreases from 50oC to 40oC)
Time taken is -kt ln e = ln T(t)−Ts / To−Ts
– (0.092) t = ln 45−25 / 70−25
– 0.092 t = -0.597
t = −0.597 / −0.092
= 6.489 min.
The Newton’s law of cooling is given by,
dT/dt = k(Tt – Ts)
Where
Tt = temperature at time t and
Ts = temperature of the surrounding,
k = constant.
The Newton’s Law of Cooling Formula is expressed by
T(t) = Ts + (T0 – Ts) e-Kt
Where,
t = time,
T(t) = temperature of the given body at time t,
Ts = surrounding temperature,
To = initial temperature of the body,
k = constant.
Please note that greater the difference in temperature between the system and surrounding, more rapidly the body temperature changes.
Example 1
The oil is heated to 70oC. It cools to 50oC after 6 minutes. Calculate the time taken by the oil to cool from 50oC to 40oC given the Surrounding temperature Ts = 25oC.
Solution:
Given:
Temperature of oil after 10 min = 50oC,
Ts = 25oC,
To = 70oC,
t = 6 min
The Newton’s law of cooling formula is expressed as
T(t) = Ts + (To – Ts) e-kt
T(t)−Ts / To−Ts = e-kt
-kt ln = ln T(t)−Ts / To−Ts
-kt = ln 50−25 / 70−25
= ln 0.555
k = – (-0.555 / 6 )
= 0.092
If Tt = 45oC (average temperature as the temperature decreases from 50oC to 40oC)
Time taken is -kt ln e = ln T(t)−Ts / To−Ts
– (0.092) t = ln 45−25 / 70−25
– 0.092 t = -0.597
t = −0.597 / −0.092
= 6.489 min.
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