Newton\'s Law of Cooling says that the rate of change of temperature of an objec
ID: 3281635 • Letter: N
Question
Newton's Law of Cooling says that the rate of change of temperature of an object is directly
proportional to the temperature difference between the object and the surrounding temperature.
Suppose that an engine part is being heat treated for toughness and hardness. It is removed from
the oven at a temperature of 180°C, and allowed to cool in a large room of temperature 20°C. After
30 minutes the temperature is 150°. Let the temperature at time t be T(t).
(a) Define the temperature difference, D= T-20, and state the given conditions in terms of this
temperature difference. In other words, give D(0), D(30). You may assume the equation
dD /dt=kD
(b) Solve the ODE and use the conditions to find any unknown constants. Rearrange it to give the
actual temperature of the object at time t.
(c) Give the temperature of the object after 2 hours.
(d) At what time do you expect the temperature to fall to 50°C
Explanation / Answer
Note : All temperatures are in degree celcius
dT/dt = k(T-Ts )
Given T(0) = 180 , Ts = 20 , and T(30) = 150
dT/dt = k(T-Ts )
T - Ts = D
D(0) = T(0) - Ts = 180 -20 = 160
D(30) = T(30) - Ts = 150 - 20 = 130
=> dT/dt = kdD/dt
dT/dt = k(T-Ts ) => dD/dt = kD
=> dD/D = kdt
Integrate both sides , with limits of D going from D(0) to D(t) and t from t = 0 to t = t
ln(D/D(0)) = k(t-0)
=> D = D(0)*ekt
Also D(30) = 130 and D(0) = 160
=> 130 = 160ek*30
=> k = 1/30 ln(130/160) = -0.0069
Therefore D = D(0)*ekt
=> T - Ts = 160 e-0.0069 t
=> T = 20+ 160 e-0.0069 t
t = 2hours = 2*60 = 120 minutes
T = 20+ 160 e-0.0069*120= 89.90
Now ,
Given T = 50 , we have to find t
=> 50 = 20 + 160 e-0.0069*t
=> t = ln ((50-20)/160) / -0.0069
=> t = 242.6 minutes