## iheartfood You place a cup of 210 degrees F coffee on a table in a room that is 68 degrees F, and 10 minutes later, it is 200 degrees F. Approximately how long will it be before the coffee is 180 degrees F? Use Newton's law of cooling: one year ago one year ago

1. iheartfood

2. iheartfood

A. 1 hour B. 35 minutes C. 15 minutes D. 45 minutes

3. satellite73

work with the difference in the temperature

4. iheartfood

idk how to do that... can u pls show me??

5. satellite73

the initial difference is $$210-68=142$$ after 10 minutes the difference is $$200-68=132$$ one way to model this is (probably not what your math teacher had in mind $T=142\left(\frac{132}{142}\right)^{\frac{t}{10}}$

6. satellite73

you want to know when it is 180 degrees, or when the temperature difference is $$180-68=112$$ so set $112=142\left(\frac{132}{142}\right)^{\frac{t}{10}}$ and solve for $$t$$

7. satellite73

you can also use $T=142e^{kt}$ but that requires solving for $$k$$

8. iheartfood

kk but idk how to solve for t since its in exponent form :(

9. satellite73

i get $\frac{112}{142}=\left(\frac{66}{71}\right)^{\frac{t}{10}}$ $\frac{\ln(\frac{112}{142})}{\ln(\frac{66}{71})}=\frac{t}{10}$ $t=\frac{10\ln(\frac{112}{142})}{\ln(\frac{66}{71})}$

10. satellite73

requires change of base

11. satellite73

oh i didn't open the link, you are supposed to do it the hard way, solve for $$k$$ first

12. iheartfood

so i get 32.50 ?? so my answer is B. 35 minutes ???

13. iheartfood

since i round ?

14. satellite73

$132=142e^{10k}$ $\frac{132}{142}=e^{10k}$ $\frac{\ln(\frac{132}{142})}{10}=k$

15. iheartfood

so k=-.0073025135?

16. satellite73

i got 32.5 as well remember to work with the temperature differences, not the actual temperature

17. iheartfood

ok so how do i know what my answer is based off of that??

18. jim_thompson5910

T(t) = Ta + (To - Ta)e^(-kt) T(t) = Ta + (210 - Ta)e^(-kt) ... Plug in To = 210 (this is the temp of the object) T(t) = 68 + (210 - 68)e^(-kt) ... plug in Ta = 68 T(10) = 68 + (210 - 68)e^(-k*10) ... Plug in t = 10 200 = 68 + (210 - 68)e^(-k*10) ... Plug in T(10) = 200 (this is the temp of the object at 10 min) Now solve for k 200 = 68 + (210 - 68)e^(-k*10) 200 = 68 + (142)e^(-10k) 200-68 = (142)e^(-10k) 132 = (142)e^(-10k) 132/142 = e^(-10k) 0.929577 = e^(-10k) e^(-10k) = 0.929577 -10k = ln(0.929577) -10k = -0.0730256 k = -0.0730256/(-10) k = 0.00730256 So your function goes from T(t) = 68 + (210 - 68)e^(-kt) to T(t) = 68 + (210 - 68)e^(-0.00730256t) ---------------------------------------------------------------- Now plug in T(t) = 180 and solve for t T(t) = 68 + (210 - 68)e^(-0.00730256t) 180 = 68 + (210 - 68)e^(-0.00730256t) 180 - 68 = (210 - 68)e^(-0.00730256t) 112 = (210 - 68)e^(-0.00730256t) 112 = (142)e^(-0.00730256t) 112/142 = e^(-0.00730256t) 0.78873239 = e^(-0.00730256t) ln(0.78873239) = -0.00730256t -0.23732819 = -0.00730256t -0.23732819/(-0.00730256) = t 32.4993139 = t t = 32.4993139 So it will take about 32.4993139 minutes for the cup to cool to 180 degrees F

19. iheartfood

oh wow thanks!! :) but my answer choices are these: A. 1 hour B. 35 minutes C. 15 minutes D. 45 minutes so do i round up and get B. 35 min as my answer??

20. iheartfood

@jim_thompson5910 @satellite73 is B my answer??

21. iheartfood

so is 35 minutes my answer?? :O

22. jim_thompson5910

maybe they're thinking in 5 min intervals

23. jim_thompson5910

because the answer is 32.4993139 minutes

24. iheartfood

idk... so what is the answer then??

25. iheartfood

so probs B. 35 minutes then??

26. jim_thompson5910

yeah probably

27. iheartfood

kk thanks!

28. iheartfood

kk thanks!