“The rate of change (decrement) of the temperature of a body is proportional to the difference between the temperature of the body and that of its surrounding medium”.
Let \( T \) be the temperature of a body at time \( t \) and \( T_0 \) be the temperature of its surrounding medium (usually air).
By Newton’s Law of Cooling
\[ \frac{dT}{dt} \propto (T - T_0) \]
i.e.,
\[ \frac{dT}{dt} = -k(T - T_0) \]
where \( k \) is a proportionality constant.
Rewriting,
\[ \frac{dT}{T - T_0} = -k dt \]
Integrating, we get
\[ \ln(T - T_0) = -kt + c_1 \Rightarrow T - T_0 = e^{-kt + c_1} \Rightarrow T = T_0 + e^{c_1} e^{-kt} \]
Thus,
\[ T = T_0 + ce^{-kt} \]
which gives the temperature of the body at any time \( t \).
1. Suppose that an object is heated to \( 300^{0}F \) and allowed to cool in a room whose air temperature is \( 80^{0}F \). If after 10 minutes the temperature of the object is \( 250^{0}F \), what will be the temperature after 20 min?.
Solution: Let \( T \) be the temperature of an object at time \( t \) and \( T_0 \) be the temperature of its surrounding medium (air). Then, by Newton’s law of cooling:
\[ T = T_0 + ce^{-kt} \quad \cdots (1) \]
Given \( T = 300^{0}F \) when \( t = 0 \) min and \( T_0 = 80^{0}F \).
From (1) we get,
\[ 300 = 80 + ce^0 \Rightarrow c = 220 \]
Also given \( T = 250^{0}F \) after \( t = 10 \) min.
From (1), we get,
\[ 250 = 80 + 220e^{-10k} \Rightarrow e^{-10k} = \frac{17}{22} \]
\[ \Rightarrow -10k = \ln \left(\frac{17}{22} \right) \Rightarrow -k = \frac{1}{10} \ln \left(\frac{17}{22} \right) \]
If \( t = 20 \) min then \( T \) =?
From (1) we get,
\[ T = 80 + 220e^{20 \left(\frac{1}{10} \ln \left(\frac{17}{22} \right) \right)} \]
\[ \Rightarrow T = 211.36^{0}F \]
2. A cake is removed from an oven at \( 210^{0}F \) and is placed in a room at constant temperature \( 70^{0}F \). After 30 minutes, the temperature of the cake is \( 150^{0}F \). When will the temperature of the cake be \( 120^{0}F \)?
Solution:Let \( T \) be the temperature of an object at time \( t \) and \( T_0 \) be the temperature of its surrounding medium (air). Then, by Newton’s law of cooling:
\[ T = T_0 + ce^{-kt} \quad \cdots (1) \]
Given \( T = 210^{0}F \) when \( t = 0 \) min and \( T_0 = 70^{0}F \).
From (1) we get,
\[ 210 = 70 + ce^0 \Rightarrow c = 140 \]
Also given \( T = 150^{0}F \) after \( t = 30 \) min.
From (1) we get,
\[ 150 = 70 + 140e^{-k(30)} \]
\[ \Rightarrow e^{-k(30)} = \frac{4}{7} \]
\[ \Rightarrow -30k = \ln \left(\frac{4}{7} \right) \Rightarrow -k = \frac{1}{30} \ln \left(\frac{4}{7} \right) \]
If \( T = 120^{0}F \) then \( t \) = ?
From (1) we get,
\[ 120 = 70 + 140e^{t/30 \ln \left(\frac{4}{7} \right)} \]
\[ \Rightarrow \frac{50}{140} = e^{t/30 \ln \left(\frac{4}{7} \right)} \]
\[ \Rightarrow \ln \left(\frac{5}{14} \right) = \frac{t}{30} \ln \left(\frac{4}{7} \right) \]
\[ t= \frac{30 \ln \left(\frac{5}{14} \right)}{\ln \left(\frac{4}{7} \right) } \]
\[ \Rightarrow t = 55.20 \text{ min} \]
Model Problems
Solve the following problmes:
1) A body is originally at \( 80^{0}C \), cools down to \( 60^{0}C \) in 20 min, the temperature of the air being \( 40^{0}C \). What will be the temperature of the body after 40 min from the original?
2) If the temperature of the air is \( 30^{0}C \) and the substance cools from \( 100^{0}C \) to \( 70^{0}C \) in 15 minutes, find when the temperature will be \( 40^{0}C \)?
3) A copper ball is heated to temperature of \( 100^{0}C \) then at time \( t = 0 \) min, it is placed in water which is maintained at a temperature \( 30^{0}C \). At the end of 3 minutes the temperature of ball is reduced to \( 70^{0}C \). Find the time at which temperature of ball drops to \( 31^{0}C \).
4) A body is heated to \( 110^{0}C \) and placed in air at \( 10^{0}C \). After 1 hour its temperature is \( 60^{0}C \). How much additional time is required for it to cool to \( 30^{0}C \)?
Answers
1) 50\(^{0} \)C
2) 52.15 min
3) 22.78 min
4) 1.32 hours