Question

Suppose you cool a pot of soup in a 20°C room. Right when you take the soup off the stove, you measure its temperature to be 180°C. After 20 minutes, the soup has cooled to 100°C. Suppose you can eat the soup when it is 80°C. How long will it take to cool to this temperature? What will be the temperature of the soup after 45 minutes? ​

Answer 1

According to the information, it can be inferred that the soup cools 5°C every minute, this means that after 45 minutes it will be at a temperature of 20°C.

How to calculate how long it takes for the soup to cool down to 80°C?

To calculate the time it takes for the soup to reach 80°C, we must take as a reference the loss in temperature that it had during 20 minutes, in which it lost 80°C, that is, it lost 4°C per minute.

According to the above, it will require 5 more minutes to reach a temperature of 80°C.

• 20 minutes ÷ 4°C = 5 minutes.

On the other hand, since the environment is at 20°C, the soup will not decrease in temperature below that temperature.

Learn more about temperature in: brainly.com/question/11464844

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Answer 2

Answer:

• 28.3 minutes
• 53.6°C

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.

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application

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

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time to 80°C

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.

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temp at 45 minutes

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.