Answer:
T ambient = 10 degrees
Explanation:
Using Newton's Law of Cooling:
T(t) = Tamb + (Ti - Tamb)*e^(-kt) ..... Eq 1
Ti = 100
We have two points to evaluate the above equation as follows:
T = 70 @ t = 10 using Eq 1
70 = Tamb + (100 - Tamb)*e^(-10k) ... Eq 2
T = 50 @ t = 20 using Eq 1
50 = Tamb + (100 - Tamb)*e^(-20k) ... Eq 3
Solving the above Eq 2 and Eq 3 simultaneously:
Using Eq 2:
(70 - Tamb) / (100 - Tamb) = e^(-10k)
Squaring both sides we get:
((70 - Tamb) / (100 - Tamb))^2 = e^(-20k) .... Eq 4
Substitute Eq 4 into Eq 3
50 = Tamb + (100 - Tamb)*((70 - Tamb) / (100 - Tamb))^2
After simplification:
50 = (Tamb (100-Tamb) + (70-Tamb)^2) / (100 - Tamb)
5000 - 50*Tamb = 4900 - 40*Tamb
Tamb = 100 / 10 = 10 degrees
