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A bowl of soup at 95!C (too hot) is placed in a 22!C room. One minute later, the soup has cooled to 83!C. When will the temperature be 49!C (just right)?

Respuesta :

someone2417

Answer:

in 3 minutes and 50 seconds

Step-by-step explanation:

soup = 95c

after 1 minute = 83c

after 2 minutes = 71c

after 3 minutes = 59

soup cools down 0.2c in 1 seconds

so 10 seconds = 2c

in 3 minutes and 50 seconds = 49c

The temperature 49°C of soup bowl will be after 5.54 minutes

Newton law of cooling

  • According to Newton's law of cooling, the rate of loss of heat from a body is directly proportional to the difference in the temperature of the body and its surroundings
  • The formula is as follow:

[tex](T-T_{s} )=(T_{0} -T_{s} )e^{-kt}[/tex]

What are the steps to solve this problem?

  • Given

[tex]T =[/tex] 83°C and 49°C

[tex]T_{s} =[/tex] 22°C

[tex]T_{0} =[/tex] 95°C

  • We have to find value of t for that first we will find value of k and applying all values in formula we get,

(83-22)=(95-22)[tex]e^{-k}[/tex]

∴ k = -ln (61/73)

Now,

(49-22) = [tex]e^{ln(61/73)t}[/tex]

∴ t = [tex]\frac{ln (27/73)}{ln(61/73)}[/tex]

∴ t = 5.54 minutes

So, after 5.54 minutes temperature will be 49°C

Learn more about Newton's Law of Cooling here:

https://brainly.com/question/13748261

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