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The temperature of a cup of coffee varies according to Newton's Law of Cooling: dT dt equals negative k times the quantity T minus A, where T is the temperature of the coffee, A is the room temperature, and k is a positive constant. If the coffee cools from 100°C to 90°C in 1 minute at a room temperature of 25°C, find the temperature, to the nearest degree Celsius of the coffee after 4 minutes.

74
67
60
42

Respuesta :

Answer:

Explanation:

Writing out the Newton's Law pf Cooling:

dT/dt = -k * (T - A),

where T is the temperature of the coffee, A is the room temperature, and k is a positive constant.

If the coffee cools from 100°C to 90°C in 1 minute at a room temperature of 25°C,

T = 100

A = 25

dT = 100 - 90 = 10

dt = 1

Putting the figures into the equation:

10/1 = -k * (100 - 25)

k = -10/75°C

After 4 minutes, dT/4 = 10/75 (100 - 25) = 10

dT = 40

Temperature after 4 minutes = 100 - 40 = 60°C

We can solve it mathematically also in simple way.

  • Common difference=100-90=-10°C=d
  • Initial=a=100

[tex]\\ \rm\Rrightarrow a_n=a+(n-1)d[/tex]

[tex]\\ \rm\Rrightarrow a_5=100+4(-10)[/tex]

[tex]\\ \rm\Rrightarrow a_5=100-40[/tex]

[tex]\\ \rm\Rrightarrow a_5=60°C[/tex]