Differential equations using Newtons law of cooling

Question

An object is placed in a room where the temperature is 20 ^◦C. The temperature of the object drops by 3 ^◦C in 4 minutes and by 5 ^◦C in 8 minutes. What was the temperature of the object when it was initially placed in the room?

Answer 1

By Newton's Law of Cooling, the temperature change is

T₀ – T₁ = T₀ – E – (T₀ – E)exp(–rt₁) and T₀ – T₂ = T₀ – E – (T₀ – E)exp(–rt₂)

where E = 20ºC is the environmental temperature. Eliminate the unknown transfer rate

r = –ln(1 – (T₀ – T₁)/(T₀ – E))/t₁ = –ln(1 – (T₀ – T₂)/(T₀ – E))/t₂, then plug in the values:

ln(1 – 3/(T₀ – E))/4 = ln(1 – 5/(T₀ – E))/8 i.e.,

1 – 5/(T₀ – E) = (1 – 3/(T₀ – E))², a quadratic equation that can be solved for 1/(T₀ – E).