Q. 70 - Calculus

Question

A cold drink is heating up from an initial temperature T(0) = 2◦C to room temperature of 22◦C according to Newton’s Law of Heating with constant of proportionality 0.05◦C. (a) Set up a differential equation describing $\frac{dT}{dt}$, and solve it to get a formula for the temperature of the drink after t minutes. (b) Use the differential equation and or it s solution to determine the units of the constant of proportionality. (c) How long will it take for the drink to warm up to within 1 degree of room temperature? 

Step-by-Step Solution

Answer

(a) $\mathrm{T}\left(\mathrm{t}\right)=22-20{\mathrm{e}}^{-0.05\mathrm{t}}$

(b) The unit of the constant of proportionality is degree centigrade. 

(c) The drink's temperature will be 21°C in 59.9 minutes. 

Step 1. Given

A cold drink is heating up from an initial temperature T(0) = 2◦C to room temperature of 22◦C according to Newton’s Law of Heating with constant of proportionality 0.05◦C. 

Part(a) Step 2. Calculation

$$\begin{gathered} \left(a\right) \mathrm{Recall} \mathrm{that} \mathrm{the} \mathrm{differential} \mathrm{equation} \mathrm{modelling} \mathrm{Newton}\text{'}\mathrm{s} \mathrm{Law} \mathrm{of} \mathrm{Cooling} \mathrm{and} \mathrm{Heating} \mathrm{for} \mathrm{ambient} \mathrm{temperature} \mathrm{A} \mathrm{is} \mathrm{given} \mathrm{by} \\ \frac{dT}{dt}=k(A-T) \left(1\right) \\ \mathrm{As} \mathrm{per} \mathrm{given} \mathrm{information} \mathrm{in} \mathrm{the} \mathrm{problem} 4 = 22 ° \mathrm{C} =0.05°\mathrm{C} \mathrm{Use} \mathrm{this} \mathrm{information} \mathrm{in} \mathrm{equation} \left(1\right) \\ \mathrm{and} \mathrm{write} \mathrm{down} \mathrm{the} \mathrm{differential} \mathrm{equation} \mathrm{representing} \mathrm{the} \mathrm{problem} \mathrm{as} \\ \frac{dT}{dt}=0.05(22-T) \\ \mathrm{Use} \mathrm{the} \mathrm{initial} \mathrm{condition}, \mathrm{namely} \mathrm{T}\left(0\right) = 2 \mathrm{to} \mathrm{write} \mathrm{down} \mathrm{the} \mathrm{initial}-\mathrm{value} \mathrm{problem} \mathrm{describing} \mathrm{the} \mathrm{given} \mathrm{situation} \mathrm{as} \\ \frac{dT}{dt}=0.05(22-T) ... T\left(0\right) = 2 \\ \mathrm{Now}, \mathrm{proceed} \mathrm{to} \mathrm{solve} \mathrm{the} \mathrm{differential} \mathrm{equation} \mathrm{in} \left(2\right). \\ \mathrm{Observe} \mathrm{that} \mathrm{the} \mathrm{differential} \mathrm{equation} \mathrm{does} \mathrm{not} \mathrm{contain} \mathrm{the} \mathrm{independent} \mathrm{variable} \mathrm{at} \mathrm{all}, \mathrm{so} \mathrm{solve} \mathrm{it} \mathrm{by} \mathrm{using} \mathrm{the} \mathrm{method} \mathrm{of} \mathrm{antidifferentiating}. \\ \int \frac{\mathrm{dT}}{22-\mathrm{T}}=0.05\int \mathrm{dt} \\ -\ln \left|22-\mathrm{t}\right|=-0.05\mathrm{t}+\mathrm{C} \\ -\ln \left|22-\mathrm{t}\right|=-0.05\mathrm{t}-\mathrm{C} \\ 22-\mathrm{t}={\mathrm{e}}^{-0.05\mathrm{t}} \\ 2=22-\mathrm{A} \\ \mathrm{A}=20 \\ \mathrm{T}\left(\mathrm{t}\right)=22-20{\mathrm{e}}^{-0.05\mathrm{t}} \end{gathered}$$

Part(b) Step 3. Calculation

Consider the differential equation representing the process of heating according to Newton's law of Heating. The phenomenon of heating is a natural process, wherein an object at a lower temperature kept in an environment with higher temperature, absorbs heat from environment and gets warmer with time. So, the constant of proportionality behaves like natural growth constant; and the growth is in temperature. Therefore, the unit of the constant of proportionality is degree centigrade.

Part (c) Step 4. Find t

In order to find the time taken by the drink to attain a temperature within 1ºof room

temperature, take T(1)=21°C in equation (3)

$21=22-20e^{-0.05t}$

Simplify the above relation and find the value of t as

$$\begin{gathered} e^{0.05t}=20 \\ t=\frac{1}{0.05}\ln 20 \end{gathered}$$

= 59.9

So, the drink's temperature will be 21°C in 59.9 minutes. Hence, the temperature will be within 1 degree of room temperature after 1 hour.