Question

A 210 degree cup of coffee is placed on a table in a climate-controlled room with the temperature set at a constant 73 degrees. After 6 minutes, the temperature of the coffee had dropped to 150 degrees. Find a function that outputs the temperature of the coffee t minutes after it is placed on the table.

Answer 1

From Newton's law of cooling:

T(t) = Ta + A*e^(k*t) eq(1)

Ta = Atmospheric temperature = temperature of room = 73 degC

at t = 0,

T0 = initial temperature of coffee = 210 degC

So,by using above values in eq (1),

210 = 73 + A*e^(k*0)

A = 210 - 73

A = 137

Also given that

at t = 6 min, T(6) = 150 C

then, 150 = 73 + 137*e^(k*6)

e^6k = (150 - 73)/137

6k = ln (77/137)

k = (1/6)*ln (77/137) = -0.096

So,function that outputs the temperature of the coffee t minutes after it is placed on the table will be:

T(t) = 73 + 137*e^(-0.096*t)

Let me know if you have any query.