At 2 pm a hot coal was pulled out of a furnace and allowed t

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 375°F, and after 20 minutes, its temperature was 315°F, find the following. (a) The temperature of the furnace. °F (b) The time when the temperature of the coal was 105°F. (Round your answer to the nearest minute.) : p.m.

Solution

Considering the rate of cooling is linear with time

    rate of cooling R = Ti- Tf / time = 375°F-315°F / 20min = 60°F/20°min = 3°F / 1 min

a) Temperature when the coal was just removed from furnace is the furnace temperature

                                 = 10 min * 3° F / 1 min + 375° F = 405°F

b) time to cool down to 105 °F at the rate of 3°F / 1 min is

                     T = 405°F - 105°F / 3 = 100 min = 3:40 PM