Question

An object is taken from a freezer at negative 7 degrees Upper C. Let t be the time in hours after the object was taken from the freezer. At time t the average temperature of the object is increasing at the rate of Upper T prime (t )equals 10 e Superscript negative 0.5 t degrees Celsius per hour. Find the temperature of the object at time t.

Answer 1

`T(t)=-20e^(-0.5t)+13`

Explanation:

The object is taken from a freezer at
`-7^0C`

`T'(t)=10e^(-0.5t)`

To determine the temperature T(t), at any time t, we take the integral of the average increase in temperature.

`\int T'(t) dt=\int (10e^(-0.5t))dt\\T(t)=-20e^(-0.5t)+k, $where k is a constant of integration$\\At \:t=0, T(t)=-7^0C\\-7=-20e^(-0.5(0))+k\\k=-7+20=13`

The temperature of the object at time t is given as:

`T(t)=-20e^(-0.5t)+13`