**Answer:**

**Step-by-step explanation:**

The **temperature** of the soup is described by **Newton's Law of Cooling**, which says the **rate of change** of temperature is **proportional to the difference** between the **temperatures** of the soup and the room. The solution to this differential equation is the **exponential function** ...

f(t) = a +b·c^(t/τ)

where a is the room temperature, b is the initial difference in temperature, and c is the fractional change in the difference in temperature over time period τ.

__

<h3>application</h3>

For the given scenario, we find ...

a = room temperature = 20 . . . . degrees C

b = (180 -20) = 160 . . . . degrees C

c = (100 -20)/(180 -20) = 80/160 = 1/2 . . . . in τ = 20 minutes

So, the formula for the temperature of the soup is ...

f(t) = 20 +160(1/2)^(t/20) . . . . . . . degrees C after t minutes

__

<h3>time to 80°C</h3>

Solving for t when f(t) = 80, we find ...

80 = 20 +160(1/2)^(t/20)

3/8 = (1/2)^(t/20) . . . . . subtract 20, divide by 160

20×log(3/8)/log(1/2) = t ≈ 28.3 . . . take logarithms, divide by coefficient of t

**It will take about 28.3 minutes to cool to 80°C**.

__

<h3>temp at 45 minutes</h3>

The temperature after 45 minutes is ...

f(45) = 20 +160(1/2)^(45/20)

f(45) ≈ 53.6 . . . . degrees C

**After 45 minutes the temperature of the soup will be about 53.6°C**.