A bowl of soup at 170 F is placed in a room of constant temp

A bowl of soup at 170° F. is placed in a room of constant temperature of 70° F. The temperature T of the soup t minutes after it is placed in the room is given by T(t ) = 70 + 100 e ^– 0.075 t - after 6 minutes

Solution

The temperature T of the soup t minutes after it is placed in the room is given by T(t ) = 70 + 100 e^-0.075t. When t=6,the temperature of the soup 6 minutes after it is placed in the room will be 70+100 e^-0.075*6 = 70 + 100e^-0.45 = 70 +100* 0.637628152 = 70 +63.7628152 = 133.7628152 , say 133.76⁰ F ( on rounding off to 2 decimal places).