Question

Group Work 2.6 Newton's Law of Cooling states that an object cools at a rate proportional to the difference between the temperature of the object and the room temperature. The temperature of the object at a time t is given by a function f(t)= alpha x^ prime prime +a where a = mc * pi temperature c = initial difference in temperature between the object and the room r = constant determined by data in the problem Suppose you make yourself a cup of tea. Initially the water has a temperature of 95 degrees * C ; 5 minutes later the tea has cooled to 65 degrees * C Problem: When will the tea reach a drinkable temperature of 40°C? Hint: Assume that the room temperature a = 22 degrees * C . First solve for r and then find tapplying the natural logarithm.​

Answer 1

Answer:

ΔT = ΔT0 e^-K T

As I understand Newton's Law of Cooling

ΔT at any time is the difference between the temperature and the surroundings

Originally       ΔT0 = 95 - 22   difference between 95 and room temperature

65 - 22 = 33 = 73 e^-KT      where t is time to cool to 65 deg

ln (33/73) = -KT       K = .794 / 5 = .159    where 5 is time to cool to 65 deg

40 - 22 = 73 e^-.159 T      where t is time to cool to 40 deg

18 = 73 e^-.159 T

ln (18 / 73) = -.159 T

T = 8.8 min

It would take 8.8 min for the object to cool to 40 deg C

Suppose the object cooled from 95 to 90 deg, then

ln 68 / 73 = -.159 T     and T = .45 min

Answer 2

Answer:

  13.2 minutes

Step-by-step explanation:

We cannot tell what your equation is supposed to be. Usually, the equation will have the form ...

  f(t) = c·e^(kt) +r

where c is the initial temperature difference (95 -22 = 73) and r is the room temperature (22). The value of k can be found from the given intermediate temperature and time.

  f(5) = 65 = 73·e^(k·5) +22

  43/73 = e^(5k)

Taking the natural log gives ...

  5k = ln(43/73)

  k = ln(43/73)/5 ≈ -0.105852

__

We want to find t for f(t) = 40. Then ...

  f(t) = 40 = 73·e^(-0.105852t) +22

  18/73 = e^(-0.105852t)

  t = ln(18/73)/-0.105852 ≈ 13.227

The tea will be drinkable after 13.2 minutes.

__

In the attached, we have used exponential regression to find the equation of the temperature curve.