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Temperature word problem

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by msquare » Mon Oct 06, 2008 1:50 pm
Can anyone explain this problem to me?

the temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees fahrenheit. if the temperature f of the coffee t minutes after it was poured can be determined by the formula f=120(2^-at) + 60, where f is in degrees fahrenheit and a is a constant, then the temperature of the coffee 30 minutes after it was poured was how many degrees faherenheit?

thanks a lot.
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Re: Temperature word problem

by Morgoth » Mon Oct 06, 2008 4:33 pm
msquare wrote:Can anyone explain this problem to me?

the temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees fahrenheit. if the temperature f of the coffee t minutes after it was poured can be determined by the formula f=120(2^-at) + 60, where f is in degrees fahrenheit and a is a constant, then the temperature of the coffee 30 minutes after it was poured was how many degrees faherenheit?

thanks a lot.
Insert the value of f=120 and t=10 to find the value of constant

f=120(2^-at) + 60

120 = 120(2^-a10) + 60

120-60 = 120(2^-a10)

60/120 = (2^-a10)

1/2 = 1/2^a10

a10 = 1
a = 1/10

now substitute the value of t=30 & a=1/10 to find f

f=120(2^-30/10) + 60

f = 120 * 1/2^3 + 60

f = 120/8 + 60

f = 15+60 = 75

Hope this helps.
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by msquare » Mon Oct 06, 2008 4:42 pm
Thank you.
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by mjoeb » Fri Apr 17, 2009 9:25 pm
hi, can you please explain how you went from:

60/120 = (2^-a10)


to

1/2 = 1/2^a10


???? i'm really confused.
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Re: Temperature word problem

by sanju09 » Sat Apr 18, 2009 2:12 am
msquare wrote:Can anyone explain this problem to me?

the temperature of a certain cup of coffee 10 minutes after it was poured was 120 degrees fahrenheit. if the temperature f of the coffee t minutes after it was poured can be determined by the formula f=120(2^-at) + 60, where f is in degrees fahrenheit and a is a constant, then the temperature of the coffee 30 minutes after it was poured was how many degrees faherenheit?

thanks a lot.
Let’s put the available values in the given relation to reveal more of it, putting values we get:

120 = 120 (2^-a*10) + 60
» 2 = 2 (2^-a*10) + 1
» 2 (2^-a*10) = 1
» (2^-a*10) = ½
» (2^-a*10) = 2^-1
» -a*10 = -1
» a = 1/10.

So the modified relation is f = 120 (2^-t/10) + 60, and when t = 30, f = 120 (2^-3) + 60 = 15 + 60 = 75 º Fahrenheit.
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by ketkoag » Mon Apr 20, 2009 10:42 am
agree with the explanation, find a and then calculate f.
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