The temperature, H, in °F, of a cup of coffee t hours after it is set out to cool is given by the following equation. H = 70 + 120(1/4)t (a) What is the coffee's temperature initially (that is, at time t = 0)? 190 °F What is the coffee's temperature after 1 hour? 100 °F What is the coffee's temperature after 2 hours? (Round your answer to one decimal place.) 2 °F (b) How long does it take the coffee to cool down to 85°F? (Round your answer to three decimal places.) 5 hr How long does it take the coffee to cool down to 75°F? (Round your answer to three decimal places.) 5 hr

Answers

Answer 1
Answer:

Answer:

The temperature a t = 0 is 190 °F

The temperature a t = 1 is 100 °F

The temperature a t = 2 is 77.5 °F

It takes 1.5 hours to take the coffee to cool down to 85°F

It takes 2.293 hours to take the coffee to cool down to 75°F

Step-by-step explanation:

We know that the temperature in °F, of a cup of coffee t hours after it is set out to cool is given by the following equation:

H(t)=70+120((1)/(4))^t

a) To find the temperature a t = 0 you need to replace the time in the equation:

H(0)=70+120((1)/(4))^0\nH(0)=70+120\cdot 1\nH(0) = 70+120\nH(0)=190 \:\°F

b) To find the temperature after 1 hour you need to:

H(1)=70+120((1)/(4))^1\nH(1)=70+120((1)/(4))\nH(1) = 70+30\nH(1)=100 \:\°F

c) To find the temperature after 2 hours you need to:

H(2)=70+120((1)/(4))^2\nH(2)=70+120((1)/(16))\nH(2) = 70+(15)/(2) \nH(2)=77.5 \:\°F

d) To find the time to take the coffee to cool down 85 \:\°F, you need to:

85 = 70+120((1)/(4))^t\n70+120\left((1)/(4)\right)^t=85\n70+120\left((1)/(4)\right)^t-70=85-70\n120\left((1)/(4)\right)^t=15\n(120\left((1)/(4)\right)^t)/(120)=(15)/(120)\n\left((1)/(4)\right)^t=(1)/(8)

\mathrm{If\:}f\left(x\right)=g\left(x\right)\mathrm{,\:then\:}\ln \left(f\left(x\right)\right)=\ln \left(g\left(x\right)\right)

\ln \left(\left((1)/(4)\right)^t\right)=\ln \left((1)/(8)\right)

\mathrm{Apply\:log\:rule}=\log _a\left(x^b\right)=b\cdot \log _a\left(x\right)

t\ln \left((1)/(4)\right)=\ln \left((1)/(8)\right)

t=(\ln \left((1)/(8)\right))/(\ln \left((1)/(4)\right))\nt=(3)/(2) = 1.5 \:hours

e) To find the time to take the coffee to cool down 75 \:\°F, you need to:

75=70+120\left((1)/(4)\right)^t\n70+120\left((1)/(4)\right)^t=75\n70+120\left((1)/(4)\right)^t-70=75-70\n120\left((1)/(4)\right)^t=5\n\left((1)/(4)\right)^t=(1)/(24)

\ln \left(\left((1)/(4)\right)^t\right)=\ln \left((1)/(24)\right)\nt\ln \left((1)/(4)\right)=\ln \left((1)/(24)\right)\nt=(\ln \left(24\right))/(2\ln \left(2\right)) \approx = 2.293 \:hours


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