Question

For a body with initial temperature T.(°C) and final temperature T (°C) Newton's Law of cooling states that:

.T(t) = T.exp(-k) where t is time (minutes) and k is a constant. Given that k = 0.043, find the time taken for a cup of
coffee to cool from 75 to 65°C.
Give your result to 2 d.p.

Answer 1

4.07 minutes for the cup of coffee to cool from 75°C to 65°C

Explanation:

To find the time taken for a cup of coffee to cool from 75°C to 65°C using Newton's Law of cooling, we can use the formula T(t) = T.exp(-kt), where T(t) is the temperature at time t, T is the initial temperature, and k is the constant.

Given that k = 0.043, T = 75°C, and we want to find the time it takes for the temperature to reach 65°C, we can substitute these values into the formula.

T(t) = T.exp(-kt)

65 = 75.exp(-0.043t)

To solve for t, we need to isolate the variable t. We can do this by dividing both sides of the equation by 75 and taking the natural logarithm of both sides.

(65/75) = exp(-0.043t)

ln(65/75) = -0.043t

Now, we can solve for t by dividing both sides of the equation by -0.043.

t = ln(65/75) / -0.043

Using a calculator, we can find the value of t to 2 decimal places.

t ≈ 4.07 minutes

Therefore, it will take approximately 4.07 minutes for the cup of coffee to cool from 75°C to 65°C