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.

∆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

are you like autistic or something?

xd

answer would be letter choice (a),   a ball resting on a shelf. would be another example of potential energy   >   stored energy > not moving.

a ball spinning on a finger   >   kinetic energy > moving energy

a ball rolling on the ground   >   kinetic energy   > moving energy

a   ball flying through the air   > kinetic energy   >   moving energy

hope that                                         : )