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Water at 100 is taken off the stove and allowed to cool for 10 minutes. In this 10 minutes,

the temperature decreased to 65. Given that the room temperature is 21, find the

temperature of the water in degrees Celsius after an additional 5 minutes waiting time and

on condition that the Newton’s Law of cooling is not violated.​

Respuesta :

Hania12

Answer:

∆θ/ΔT  ∝(θ-Te)

((100-65))/10  ∝(82.5-21) --------------------------------------------------------------------------(1)

((100-T))/15  ∝(((100+T))/2-21) ---------------------------------------------------------------------------(2)

(2)/(1)

T = 52.73 C

Explanation:

Decreases of temperature is proportional to excess temperature according to the Newton’s cooling law

∆θ/ΔT  ∝(θ-T1)

Here we have to use θ for average temperature from 100C to 65 C. Because temperature  is not lineally deceased. There for  

θ=((100+65))/2=82.5 C    

Considering 1 Statement,

((100-65))/10  ∝(82.5-21) --------------------------------------------------------------------------(1)

Considering 2 statement,

Assume that final temperature is T after 15 minutes time,

Then ,

θ=((100+T))/2    

((100-T))/15  ∝(((100+T))/2-21) ---------------------------------------------------------------------------(2)

(2)/(1)

T = 52.73 C

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