Freddy Friedman

2022-07-27

Suppose it is known that the coffee cools at a rate of 1degree C when its temperature is 70 degrees C. The room is 20degrees C.

a) What does the differential equation become in thiscase?

b) sketch a direction field and use it to sketch the solution curve for the initial-value problem. What is the limiting value of the temperature?

c) Use Euler's method with step size h=2 minutes to estimate the temperature of the coffee after 10 minutes

a) What does the differential equation become in thiscase?

b) sketch a direction field and use it to sketch the solution curve for the initial-value problem. What is the limiting value of the temperature?

c) Use Euler's method with step size h=2 minutes to estimate the temperature of the coffee after 10 minutes

abortargy

Beginner2022-07-28Added 19 answers

first things first you need to solve for K from the original equation:

dC/dt=-K( C-A) where A=ambient temp, C=coffee temp, Kneeds to be solved, dC/dt= rate of decline

C=70, dC/dt= -1, A= 20

-1 = -K (70-20) K = -1/50

0=70

2mins = 70 + 2(-1/50(70-20)) = 68

4mins = 68 + 2(-1/50(68-20)) = 66.08

6mins = 66.082 + 2(-1/50(66.08-20)) = 64.24

8mins = 64.24 + 2(-1/50(64.24-20)) = 62.47

10mins = 62.47 + 2(-1/50(62.47-20)) = 60.77

you can also solve it for the actual value by inverting thedC/dt=-k(C-A) to get dt/dC = -50/(C-20)

dC/dt=-K( C-A) where A=ambient temp, C=coffee temp, Kneeds to be solved, dC/dt= rate of decline

C=70, dC/dt= -1, A= 20

-1 = -K (70-20) K = -1/50

0=70

2mins = 70 + 2(-1/50(70-20)) = 68

4mins = 68 + 2(-1/50(68-20)) = 66.08

6mins = 66.082 + 2(-1/50(66.08-20)) = 64.24

8mins = 64.24 + 2(-1/50(64.24-20)) = 62.47

10mins = 62.47 + 2(-1/50(62.47-20)) = 60.77

you can also solve it for the actual value by inverting thedC/dt=-k(C-A) to get dt/dC = -50/(C-20)

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