The temperature of a cooling liquid over time can be modeled by the exponential function 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°C. What was the initial temperature?

Question

The temperature of a cooling liquid over time can be modeled by the exponential function 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°C. What was the initial temperature?

shane_b · Answer

To be clear, is the equation:$$\Large T(x)=(60)(\frac{1}{2})^\frac{x}{30}+20$$

anonymous · Answer

^yes

shane_b · Answer

- To graph it I'd just use a graphing calculator/program.
- To find out how long it takes for the temp to get to 28 degrees C you can either solve this for x:$$\Large 28=(60)(\frac{1}{2})^\frac{x}{30}+20$$or do a trace on the graph to see where y = 28.
- The last part is very simple.  Plug in x=0 (time = 0) and you'll see that the equation reduces to:$$\Large (60)(\frac{1}{2})^\frac{0}{30}+20=60+20=80C$$

shane_b · Answer

BTW...when you solve for x in the second part above you should get this: http://www.wolframalpha.com/input/?i=solve+for+x%3A+28%3D60*%281%2F2%29^%28x%2F30%29%2B20