Differential equation Newton law of cooling At 2 pm a hot co

Differential equation/ Newton law of cooling. At 2 p.m., a hot coal was pulled out of a furnace and allowed to cool at room temperature (75°F). If, after 10 minutes, the temperature of the coal was 435°F, and after 20 minutes, its temperature was 363°F, find the following.(b) The time when the temperature of the coal was 95°F. (Round your answer to the nearest minute.) I already know the T(o)=525F.

Solution

Here\' the answer to the problem:

525°F is the original temperature

After 10 min, 435° is the temperature

After 20 min, 363° is the temperature

Hence, the decrease rate of temperature is 363°-435° / 10 = 9 degrees/min

b ) The time when the coal will be 95° ?

This is a difference of 268° ( 270° approx) degrees . This will happen in 270°/9 = 30 minutes

Hence, 30 minutes is the approx time by which the temperature will go doen to 95° F