## iheartfood Group Title 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 Group Title

2. iheartfood Group Title

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

3. satellite73 Group Title

work with the difference in the temperature

4. iheartfood Group Title

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

5. satellite73 Group Title

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 Group Title

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 Group Title

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

8. iheartfood Group Title

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

9. satellite73 Group Title

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 Group Title

requires change of base

11. satellite73 Group Title

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

12. iheartfood Group Title

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

13. iheartfood Group Title

since i round ?

14. satellite73 Group Title

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

15. iheartfood Group Title

so k=-.0073025135?

16. satellite73 Group Title

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

17. iheartfood Group Title

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

18. jim_thompson5910 Group Title

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 Group Title

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 Group Title

@jim_thompson5910 @satellite73 is B my answer??

21. iheartfood Group Title

so is 35 minutes my answer?? :O

22. jim_thompson5910 Group Title

maybe they're thinking in 5 min intervals

23. jim_thompson5910 Group Title

because the answer is 32.4993139 minutes

24. iheartfood Group Title

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

25. iheartfood Group Title

so probs B. 35 minutes then??

26. jim_thompson5910 Group Title

yeah probably

27. iheartfood Group Title

kk thanks!

28. iheartfood Group Title

kk thanks!