Question

An object's temperature cools exponentially after it is removed from a furnace (hot oven). If its initial temperature is 730°C and it cools at a rate of 2.95% per minute, find the object's temperature after 18.0 minutes.

Answer 1

Answer:

The object's temperature after 18 minutes is of 426ºC.

Step-by-step explanation:

The temperature after t minutes is modeled by an exponential function, which has the following format:

$T(t) = T(0)(1-r)^{t}$

In which T(0) is the initial temperature, and r is the cooling rate, as a decimal.

Its initial temperature is 730°C and it cools at a rate of 2.95% per minute

This means that $T(0) = 730, r = 0.0295$. So

$T(t) = T(0)(1-r)^{t}$

$T(t) = 730(1-0.0295)^{t}$

$T(t) = 730(0.9705)^{t}$

Find the object's temperature after 18.0 minutes.

This is $T(18)$. So

$T(t) = 730(0.9705)^{t}$

$T(18) = 730(0.9705)^{18} = 426$

The object's temperature after 18 minutes is of 426ºC.