3. State Newton's law of cooling, write its expresion in the form of specific heat.
Answers
Answer:
What is Newton’s Law of Cooling?
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.
Definition: 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.
Solved Examples
Example 1: A body at temperature 40ºC is kept in a surrounding of constant temperature 20ºC. It is observed that its temperature falls to 35ºC in 10 minutes. Find how much more time will it take for the body to attain a temperature of 30ºC.
Solution:
From Newtons law of cooling, qf = qi e-kt
Now, for the interval in which temperature falls from 40 to 35oC.
(35 – 20) = (40 – 20) e-k.10
e-10k = 3/4
k = [ln 4/3]/10 . . . . (a)
Now, for the next interval;
(30 – 20) = (35 – 20)e-kt
e-kt = 2/3
kt = ln 3/2 . . . . (b)
From equation (a) and (b);
t = 10 × [ln(3/2)/ln(4/3)]= 14.096 min.
Aliter : (by approximate method)
For the interval in which temperature falls from 40 to 35oC
<q> = (40 + 35)/2 = 37.5ºC
From equation (4);
dθ/dt = k(<q> – q0)
(35 – 40)/10 = k(37.5 – 20)
k = 1/32 min-1
Now, for the interval in which temperature falls from 35oC to 30oC
<q> = (35 + 30)/2 = 32.5oC
From equation (4);
(30 – 35)/t = (32.5 – 20)
Therefore, the required time t = 5/12.5 × 35 = 14 min.
Example 2: 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
On substituting the given data in Newton’s law of cooling formula, we get;
T(t) = Ts + (Ts – To) 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 T(t) = 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.
Example 3: Water is heated to 80oC for 10 min. How much would be the temperature if k = 0.056 per min and the surrounding temperature is 25oC?
Solution:
Given:
Ts = 25oC,
To = 80oC,
t = 10 min,
k = 0.056
Now, substituting the above data in Newton’s law of cooling formula,
T(t) = Ts + (To – Ts) × e-kt
= 25 + (80 – 25) × e-0.56 = 25 + [55 × 0.57] = 45.6 oC
Temperature cools down from 80oC to 45.6oC after 10 min.