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Newton's law of cooling states that the temperature of an object changes at a rate proportional to the difference between its temperature and that of its surroundings. Suppose that the temperature of a cup of coffee obeys Newton's law of cooling. If the coffee has a temperature of 210 degrees Fahrenheit when freshly poured, and 2.5 minutes later has cooled to 191 degrees in a room at 64 degrees, determine when the coffee reaches a temperature of 156 degrees.

User Valery
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Answer:

Step-by-step explanation:

The problem is based on Newton's law of cooling .

According to Newton's law

dQ / dt = k ( T - T₀ ) ,

dT / dt = k' ( T - T₀ ) ; dT / dt is rate of fall of temperature.

T is average temperature of hot body , T₀ is temperature of surrounding .

In the first case rate of fall of temperature = (210 - 191) / 2.5

= 7.6 degree / s

average temperature T = (210 + 191) /2

= 200.5

Putting in the equation

7.6 = k' ( 200.5 - 64 )

k' = 7.6 / 136.5

= .055677

In the second case :---

In the second case, rate of fall of temperature = (191 - 156) / t

= 35 / t , t is time required.

average temperature T = (156 + 191) /2

= 173.5

Putting in the equation

35 / t = .05567 ( 173.5 - 64 )

t = 5.74 minute .

User BriiC
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