Ytukyg

I have been trying to solve this problem:

A body initially at $100$ degrees Celsius cools to 60 degrees Celsius in $5$ minutes and to $40$ degrees Celsius in $10$ minutes.

What is the temperature of the surroundings?

What will be the temperature in $15$ minutes?

I was able to solve the first question without using any integration and stuff. Please help me with this problem.

Start with the differential equation:

$T' = - K(T-E)$

K and E are constants representing the rate of heat transfer and the temperature of the environment.

Heat is lost to the environment in proportion to the temperature differential.

Solve the differential equation and use the information given to find $K, E$

From there you will have an equation for T as a function of time.

This can be done without using "integration and stuff" but if you don't have the intuition, I suggest you do learn the "stuff."

Without the "stuff"

How much heat is lost to the environment in each 5 minute interval? How much will be lost in the next 5 minutes? It is all moving in constant proportions, and can be solved using partial sums of a geometric series.

UPDATE

$T' = - K(T-E)\
frac {T'}{T-E} = -K\
T(t) - E = Ae^{-Kt}\
T(0) = A + E = 100\
T(5) = Ae^{-5k} + E = 60\
T(10) = Ae^{-10k} + E = 40\
T(0)-T(5) = A(1-e^{-5k}) = 40\
T(5)-T(10) = A(e^{-5k}-e^{-10k}) = 20\
2A(e^{-5k}-e^{-10k}) = A(1-e^{-5k})\
2Ae^{-5K}(1-e^{-10k}) = A(1-e^{-5k})\
2e^{-5K} = 1\
e^{-5K} = frac 12$

$T(0)-T(5) = A(1-e^{-5k}) = 40\
frac 12 A = 40\
A = 80\
A + E = 100\
E = 20$

$T(15) = 80e^{-15Kt} + 20\
80(frac 18) + 20\
30$

The intuitive approach.

Every 5 minutes the body loses half of the excess heat.

It lost 20 degrees between minute 5 and minute 10
It will loose 10 degrees between minute 10 and minute 5.

And at the limit.

$100 - 40 - 20 - 10 - 5 - frac {5}{2} cdots = 100-80 = 20 = E$