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A cup of coffee at 95 degrees Celsius is placed in a room at 25 degrees Celsius. Suppose that the coffee cools at a rate of 2 degrees Celsius per minute when the temperature of the coffee is 70 degrees. The differential equation describing this has the form

Sagot :

Answer:

See Explanation

Step-by-step explanation:

For an object at temperature T and supposing that the ambient temperature is Ta then we can write the differential equation that typifies the Newton law of cooling as follows;

dT/dt=-k(T-Tₐ)

So

dT/dt = 2 degrees Celsius per minute

T = 70 degrees Celsius

Ta = 25 degrees Celsius

2 = -k(70 - 25)

-k = 2/(70 - 25)

k = - 0.044

Hence we can write;

dT/dt=-(- 0.044)(95-25)

dT/dt= 3 degrees Celsius per minute

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