An object is placed in a room where the temperature is 20 degrees C. The temperature of the object drops by 5 degrees C in 4 minutes and by 7 degrees C in 8 minutes. What was the temperature of the object when it was initially placed in the room

28.3°C

Explanation:

Using

T(t) = (T(0) - 20)*(e^(-k*t)) + 20

for some positive number k, and some initial temperature T(0).

Boundary conditions:

T(4) = T(0) - 5 (i)

T(8) = T(0) - 7 (ii)

==> solving for T(0) and k :

(i):

(T(0) - 20)*(e^(-k*4)) + 20 = T(0) - 5 ==>

(T(0) - 20)*(e^(-k*4)) = T(0) - 20 - 5

(T(0) - 20)*(e^(-k*4)) = (T(0) - 20) - 5

5 = (T(0) - 20) - (T(0) - 20)*(e^(-k*4))

5 = (T(0) - 20) * ( 1 - e^(-k*4) )

(ii):

(T(0) - 20)*(e^(-k*8)) + 20 = T(0) - 7

(T(0) - 20)*(e^(-k*8)) = (T(0) - 20) - 7

7 = (T(0) - 20) - (T(0) - 20)*(e^(-k*8))

7 = (T(0) - 20) * (1 - e^(-k*8))

In both results, subsitute x = e^(-4k) and C = (T(0) - 20)

(i): 5 = C * (1 - x)

(ii): 7 = C * (1 - x^2) = C * (1-x)*(1+x)

Substitute C*(1-x) from (i) into (ii):

(ii): 7 = 5*(1+x) ==> (1+x) = 7/5 ==> x = 2/5

back into (i):

(i): 5 = C * (1 - 2/5) ==> 5 = C * 3/5 ==> C = 25/3

C = T(0) - 20 ==>

T(0) = C + 20 = 25/3 + 20 = 25/3 + 60/3 = 85/3

= 28.3°C

that would be b i just took the test : )

the answer is b ) square root of 5