What was the intial temperature in the problem below?Problem: T(x) = 60(1/2)^(x/30) + 20 , where T(x) is the temperature in degrees Celsius and x is the elapsed time in minutes. Graph the...
What was the intial temperature in the problem below?
Problem: T(x) = 60(1/2)^(x/30) + 20 , where T(x) is the temperature in degrees Celsius and x is the elapsed time in minutes. Graph the function and determine how long it takes for the temperature to reach 28 deg Celsius.
The temperature of a cooling liquid over time can be modeled by the exponential function:
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1 Answer
lemjay | High School Teacher | (Level 3) Senior Educator
The given equation is:
`T(x) = 60 (1/2)^(x/30) + 20`
For T = 28 deg Celsius, x = ?
Solution:
Substitute the value of T to the above equation.
`28 = 60 (1/2)^(x/30) + 20`
`28 - 20 = 60(1/2)^(x/30)`
`8 = 60(1/2)^(x/30)`
`8/60 = (1/2)^(x/30)`
`2/15 = (1/2)^(x/30)`
Use the property of logarithm `log_b m^n = n log_b m`
So, take the logarithm of both sides of the equation. For this, let's use the natural logarithm.
`ln(2/15) = ln (1/2)^(x/30)`
`ln(2/15) = (x/30) ln(1/2)`
`ln(2/15) / (ln(1/2)) = x/30`
`30 ln(2/15)/(ln(1/2)) = x`
87.2 mins = x
For the initial temperature, set x=0. Then solve for T(0).
Solution:
`T(0) = 60(1/2)^(0/30) + 20 = 60 (1/2)^0 + 20 = 60 (1) + 20 = 80 deg Celsius`
Answer: The intial temperature is 80 deg Celsius. To lower the temperature from 80 to 28 deg Celsisus, it takes 87.2 minutes.