-5 + 4 on a number line
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Answer:
Newton's law of cooling says that:
T(t) = Tₐ + (T₀ - Tₐ)*e^(k*t)
or:
in the differential form.
where:
T is the temperature as a function of time
Tₐ is the ambient temperature, in this case, 70F
T₀ is the initial temperature of the object, in this case, 150F
k is a constant, and we want to find the value of k.
Then our equation is:
T = 70F + (150F - 70F)*e^(k*t)
Now we also know that after a minute, or 60 seconds, the temperature was 135F
then:
135F = 70F + (150F - 70F)*e^(k*60s)
We can solve this for k:
135F = 70F + 80F*e^(k*60s)
135F - 70F = 80F*e^(k*60s)
65F = 80F*e^(k*60s)
(65/80) = e^(k*60s)
Now we can apply the Ln(x) function to both sides to get:
Ln(65/80) = Ln(e^(k*60s))
Ln(65/80) = k*60s
Ln(65/80)/60s = k = -0.0035 s^-1
Then the differential equation is: