Newton's Law of Cooling of coffee

[ajr246]

Homework Statement

Suppose that a cup of hot coffee at a temperature of 185⁰ is set down to cool in a room where the temperature is kept at 70⁰. What is the temperature of the coffee 10 minutes later?

Homework Equations

f(t)=(T₀ - T₁₎)e^-kt + T₁
where T₀ is the initial temperature of the coffee,
T₁ is the temperature of the room,
f(t) is the temperature of the coffee after t minutes,
k is a negative constant

The Attempt at a Solution

I am studying precalculus independently. This problem is from Cohen, Precalculus with Unit-Circle Trigonometry, 4th ed.

The problem statement doesn't seem to give enough data to solve it. I can't figure out how to find k in order to find f(t). Is there a technique I'm not seeing?

I've gotten this far:
f(10)=(185-70)e^-k10+70
f(10)=115e^-10k+70

 

[HallsofIvy]

Homework Statement

Suppose that a cup of hot coffee at a temperature of 185⁰ is set down to cool in a room where the temperature is kept at 70⁰. What is the temperature of the coffee 10 minutes later?

Homework Equations

f(t)=(T₀ - T₁₎)e^-kt + T₁
where T₀ is the initial temperature of the coffee,
T₁ is the temperature of the room,
f(t) is the temperature of the coffee after t minutes,
k is a negative constant

The Attempt at a Solution

I am studying precalculus independently. This problem is from Cohen, Precalculus with Unit-Circle Trigonometry, 4th ed.

The problem statement doesn't seem to give enough data to solve it. I can't figure out how to find k in order to find f(t). Is there a technique I'm not seeing?

I've gotten this far:
f(10)=(185-70)e^-k10+70
f(10)=115e^-10k+70

You're right. Since you do not know either k or Q you cannot solve that equation for either one. I suspect that you have copied the problem incorrectly. You have to be given the temperature initially and at some other time in order to find k. Then you can find the temperature for any t.

 

[ajr246]

Thanks.