Question

Suppose you have just poured a cup of freshly brewed coffee with temperature 95∘C in a room where the temperature is 25∘C. Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings. Therefore, the temperature of the coffee, T(t), satisfies the differential equation dTdt=k(T−Troom) where Troom=25 is the room temperature, and k is some constant. Suppose it is known that the coffee cools at a rate of 2∘C per minute when its temperature is 65∘C. A. What is the limiting value of the temperature of the coffee?B. What is the limiting value of the rate of cooling?C. Find the constant k in the differential equation.D. Use Euler's method with step size h=2 minutes to estimate the temperature of the coffee after 10 minutes.

Answer 1

Answer:

A. 25°C

B. 0°C/min

C. -0.05/min

D. 66°C

Explanation:

A. In this case, the value of the temperature of coffee will never reach exactly room temperature, but it will approach it. This value to which it approaches is called the limiting value, that is to say, 25°C.

The rate of change of temperature present for this limiting value (25°C) is indeed the limiting value of the rate of cooling. Introducing this value (T=25°C) in the equation:

$\frac{dT}{dt} =k(T-T_{room} )=k(25C-25C)=0$

B. Zero is the limiting value to which the rate of cooling will approach when the coffee temperature reaches the limiting value.

C. To find the constant k in the differential equation, we use the fact that coffee cools at a rate of 2∘C per minute when its temperature is 65∘C.

$-2C/min=k(65C-25C)$

Note that the sign of -2C/min is negative because the temperature is decreasing with time. Solving for k

$k=\frac{-2C/min}{(65C-25C)} =-0.05/min$

To estimate the temperature of the coffee after 10 minutes, we use Euler's formula. We start with ti=t0=0

$\frac{dT(t=t_{i} )}{dt}=f(t_{i} , T_{i} )$

$T_{i+1} = T_{i}+h*f(t_{i}, T_{i})$

$f(t_{0} , T_{0} )=(-0.05/min)(95C-25C)=-3.5C/min$

$T_{1}(2min) =95C-3.5C/min*2min=88C$

$f(t_1} , T_{1} )=(-0.05/min)(88C-25C)=-3.15C/min$

$T_{2}(4min) =88C-3.15C/min*2min=88C=82C$

$f(t_2} , T_{2} )=(-0.05/min)(82C-25C)=-2.9C/min$

$T_{3}(6min) =82C-2.9C/min*2min=76C$

$f(t_3} , T_{3} )=(-0.05/min)(76C-25C)=-2.6C/min$

$T_{4}(8min) =76C-2.6C/min*2min=71C$

$f(t_{4} , T_{4} )=(-0.05/min)(71C-25C)=-2.3C/min$

$T_{5}(10min) =71C-2.3C/min*2min=66C$

The temperature of the coffee after 10 minutes is 66°C.

Answer 2

The limiting temperature is 25°C and the rate of limiting temperature is 0°C.

What is Limiting temperature?

The temperature of the system can not reach the temperature of its surroundings but it approaches it. Thus, in the given case the limiting temperature is 25°C.

The limiting value of the rate of cooling:

The minimum temperature at which rate the coffee can be cooled down. Thus the value is 0°C/min.

Therefore, the limiting temperature is 25°C and the rate of limiting temperature is 0°C.

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