Question

The temperature T of a liquid which is has been placed in a refrigerator to cool down is givenby T (t)=10+90e -0.041 C, where t is the time in minutes. Include units.

A) Find the initial temperature of the liquid when time t=0 minutes.
B) Find the temperature of material after 10 minutes.
C) Find the time for the temperature to reduce to 40°C. Include units.

Answer 1

The initial temperature of the liquid is 100°C. After 10 minutes, the temperature is approximately 50.56°C. The time for the temperature to reduce to 40°C can be found by solving the equation -ln(1/3)/0.041.

Step-by-step explanation:

To find the initial temperature of the liquid when time t=0 minutes, we substitute t=0 into the equation T(t)=10+90e-0.041t. This gives us T(0)=10+90e-0.041(0)=10+90e0=10+90=100°C.

To find the temperature of the material after 10 minutes, we substitute t=10 into the equation T(t)=10+90e-0.041t. This gives us T(10)=10+90e-0.041(10)=10+90e-0.41≈10+40.56=50.56°C.

To find the time for the temperature to reduce to 40°C, we set T(t)=40 and solve for t. 40=10+90e-0.041t =>30=90e-0.041t => e-0.041t=30/90=1/3 => -0.041t=ln(1/3) => t=-ln(1/3)/0.041. We can use a calculator to find an approximate value for t.