Water left at 150 degrees to cool in 80 degree rm. After t min temp of water is given by t(t)=80+70e^-0.096t
Find temp 10 min later
Please label the School Subject. Science?
Sra
Just substitute 10 for t (minutes) and perform the indicated calculation.
T(10 min) = 80 + 70 e^-0.96 = 80 + 26.8
= 106.8 F
(I am assuing Fahrenheit temperatures, since I cannot imagine an 80 C room, or liquid water at 150 C.)
To find the temperature of the water 10 minutes later, you need to substitute t = 10 into the equation t(t) = 80 + 70e^(-0.096t).
Substituting t = 10 into the equation:
t(10) = 80 + 70e^(-0.096 * 10)
Simplifying:
t(10) = 80 + 70e^(-0.96)
Using a calculator or mathematical software, you can calculate the approximate value of e^(-0.96). Let's assume it to be approximately 0.3836.
t(10) = 80 + 70 * 0.3836
Simplifying further:
t(10) = 80 + 26.852
Calculating the final result:
t(10) ≈ 106.852
Therefore, the temperature of the water 10 minutes later is approximately 106.852 degrees Fahrenheit.