This document discusses applying Newton's Law of Cooling to solve a first order differential equation problem involving the cooling of hot coffee. The coffee was initially at 180°F and cooled to 160°F after 10 minutes in a kitchen at 70°F. Using Newton's Law, which states the cooling rate is proportional to the temperature difference between the object and surroundings, an equation relating temperature over time was derived. The equation found that the temperature at 10:15, or t=15 minutes, was 151.40°F.

1. MAT 131 : Ordinary and
Partial Differential Equation
Ms. Saba Fatema ( Sr. Lecturer )
Name:- Abir Hasnat
Id :- 161-15-6789

2. Application of 1st Order ODE
1.Population Dynamics
2.Growth and Decay Problems
3.Newtons Law of Cooking or Heating
4.Circuit Problem

3. Suppose, at 10:00am you took a cup of hot
instant coffee from your microwave oven and
placed it in a nearby kitchen counter to cool. At
this instant, the temperature of the coffee is
180℉ and 10 minutes later it was 160℉. Assume
that the constant temperature of the kitchen
was 70℉. What was the temperature of the
coffee at 10:15 am?

4. Newton's law of cooling states that the rate
of temperature loss of a body is
to the difference in temperatures between
the body and its surroundings.

5. Let 𝑦(𝑡) is the temperature of the coffee after t min and 𝑦𝑠 the temp of kitchen.
According to Newton’s Law of cooling
→
𝑑𝑦
𝑑𝑡
∝ 𝑦 − 𝑦𝑠
→
𝑑𝑦
𝑑𝑡
= 𝑘 𝑦 − 70
→
𝑑𝑦
𝑦−70
= k 𝑑𝑡
→ ln(y-70) = kt + c
→ y-70 = 𝑒 𝑘𝑡+𝑐
→ y = A𝑒 𝑘𝑡
+70
∴ y(t) = A𝑒 𝑘𝑡
+70 ……….. … (1)
Given, y(0) = 180
180 = A𝑒 𝑘.0 +70
∴ A = 110
∴ y(t) = 110𝑒 𝑘𝑡
+70 ……….. (2)
Again given that,
when, t = 10 then y = 160
∴ 160 = 110𝑒10𝑘
+70
→ 𝑒10𝑘
=
160−70
110
→ 𝑒 𝑘 = (
9
11
)
1
10

6. ∴ eqn (2) → y(t) = 110 X (
9
11
)
𝑡
10 +70 ……….. (3)
this is the required formula.
Now,
when t= 15,
y= 110 X (
9
11
)
15
10 +70
∴ y= 151.40
So, that at 10:15 am the temperature was 151.40℉.
(Ans)

7. Thank You