, where T is the temperature of the tea, A is the room temperature, and k is a positive constant. If the water cools from 100°C to 80°C in 1 minute at a room temperature of 60°C, find the temperature, to the nearest degree Celsius of the coffee after 4 minutes.
Timothy H. answered • 05/09/20
Patient and Knowledgeable SAT Math, Calculus, Geometry Tutor
Hi Holly,
Your question reminds me a lot of a differential equations class question, however I have seen this type of problem¯ answered in Calculus. That's because this is an anti-differentiation problem where we get like variables on one side of the equation and anti-differentiate to obtain the solution to the equation. Using the other information in the problem allows us to solve for the constant, which we then use to solve the expression. So, what does this look like you might ask?
Let dT/dt=-k(T-A).
The first step would be to assess what this question looks like. In particular, it has four variables: T,t,k and A. However, we are already given A and k is a constant. So, there are only really two variables: T and t.
The second step would be following it up by splitting apart the infinitesimals dT and dt and get like terms with like terms. So, dT/(T-A)=-k dt. This left side could look tricky.
The third step would be to take the integral of both sides: ∫dT/(T-A)= ∫-k dt.
The left hand side is just ln(T-A), while the right hand side is just -kt. Now, don't forget your constant!
Now, on to the fourth step would be to solve for the temperature in terms of time:
the inverse of a logarithm is the exponent. Since the logarithm's base is e, take e and raise it to ln(T-A):
e^(ln(T-A))=T-A=e^(-kt+C)=C0e^(-kt)=> T=A+C0e^(-kt).
Next, apply the condition that at 0 min, the temperature was 100°C to find the constant. Finally, after 1 min, the temperature of the coffee is 80°C to find the constant k. This should lead you to solve your problem.
