The temperature of a cup of coffee varies according to Newton's Law of Cooling: dT/dt = -k(T - A), where T is the temperature of the tea, A is the room temperature, and k is a positive constant. If the water cools from 100°C to 80°C in 1 minute at a room temperature of 30°C, find the temperature, to the nearest degree Celsius of the coffee after 4 minutes.

T(t) = 30 + 70e^(-1.25t) isn't correct answer.

It's like this.

k = ln ( 7 / 5 )

T = A + c ∙ e ^ ( - k t )

T = 30 + 70 ∙ e ^ ( - k t )

T = 30 + 70 ∙ e ^ [ - ln ( 7 / 5 ) ∙ t ]

T = 30 + 70 ∙ [ e ^ ln ( 7 / 5 ) ] ^ ( - t )

T = 30 + 70 ∙ ( 7 / 5 ) ^ ( - t )

T = 30 + 70 ∙ 1.4 ^ ( - t )

T(4) = 30 + 70 ∙ 1.4 ^ ( - 4 )

T(4) = 30 + 70 ∙ 0.26030825

T(4) = 30 + 18.2215775

T(4) = 48.2215775 °C

T(4) = 48°C

to the nearest °C

It is better to keep k value in exact form till you get final answer. So use k=ln(5/7) to find T(4).

Also ln(5/7) can be written as -ln(7/5) using properties of logs
T(4)=30+70e^(ln(7/5)*4) =48.2

The temperature, f(t) of a cup of coffee after t minutes can be determined by the equation f(t)=63(0.91)T

To find the temperature of the coffee after 4 minutes, we can use the given information about the initial temperature, room temperature, and the rate at which the coffee cools.

Let's break down the problem step by step.

Step 1: Understand the given information.
We are given that the temperature of the coffee cools from 100°C to 80°C in 1 minute at a room temperature of 30°C.

Step 2: Identify the variables and their values.
Let:
- Initial temperature of the coffee (T0) = 100°C
- Final temperature of the coffee (T) = 80°C
- Room temperature (A) = 30°C
- Rate constant (k) = unknown

Step 3: Use the given information to find the rate constant (k).
We can use the given information to determine the rate constant (k) by rearranging Newton's Law of Cooling equation and solving for k.

dT/dt = -k(T - A)

Since we know that the coffee cools from 100°C to 80°C in 1 minute, we can substitute these values into the equation:

dT/dt = -k(80 - 30)

We also know that the derivative (dT/dt) is equal to the change in temperature divided by the change in time:

dT/dt = (T - T0) / (t - t0)

Substituting the known values:

(T - T0) / (t - t0) = -k(80 - 30)

(80 - 100) / (1 - 0) = -k(80 - 30)

-20 / 1 = -k(50)

-20 = -50k

Dividing both sides of the equation by -50:

k = 20 / 50

k = 0.4

Therefore, the value of the rate constant (k) is 0.4.

Step 4: Use the rate constant to find the temperature after 4 minutes.
Now that we have the rate constant (k), we can use it to find the temperature of the coffee after 4 minutes.

We can rearrange the Newton's Law of Cooling equation and solve for T, the temperature of the coffee after 4 minutes:

dT/dt = -k(T - A)

dT = -k(T - A) dt

Integrating both sides of the equation with respect to t:

∫ dT = -k ∫ (T - A) dt

(T - A) = -k t + C1

T = -kt + C1 + A

We can now substitute the known values into the equation:

T = -0.4(4) + C1 + 30

Since we need to find the temperature to the nearest degree Celsius, we can ignore the constant term (C1) because it will be rounded off.

T = -0.4(4) + 30

T = -1.6 + 30

T = 28.4

Therefore, the temperature of the coffee after 4 minutes is approximately 28°C.

dT/dt = -k(T-A)

dT/(T-A) = -k dt
ln(T-A) = -kt + c
T-A = c e^(-kt)
T = A + c e^(-kt)
Now just plug in you data point, and you have
30 + c e^-k = 80
ce^-k = 50
Now, what you left out is that c = 100-30 = 70
e^-k = 5/7
k = ln(7/5) = 0.336
So, now you know that
T(t) = 30 + 70e^(-1.25t)
So now you can find T(4)