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If u pour a cup of coffee that is 200 degress F, and set it on a desk in a room that is 68 degrees F, & 10 minutes later it is 145 degres F, what tempetrature will it be 15 min after u originally poured it?

Aug 30th, 2015

if you lost 55 degrees in 10 min you would loose 5.5 degrees in 1 min then you multiply by 15. 15* 5.5 = 82.5 degrees then you subtract from 200. 200-82.5 = 117.5 degrees in 15 min <--------Answer

Aug 30th, 2015

i got 127 degrees, using the formula T=Ambient temp +(original temp -ambeint temp)e^-kt

Aug 30th, 2015

i got 127 degrees, using the formula T=Ambient temp +(original temp -ambeint temp)e^-kt

Aug 30th, 2015

But if you think it logically, my above calculations would be correct. Not using newton's law of cooling.

Aug 30th, 2015

can you do it using newtons law of cooling, that is how my teacher wants it to be done

Aug 30th, 2015

K let me try it using Newton's law of cooling.

Aug 30th, 2015

Hi,

So here is the step by step explanation for your question:

Newton's Cooling Law: T(a)=Ce^(-kt)+Ta

Step 1:

T(a): 68F

T(0): 200F

T(10): 145F

T(15): ?

Step 2:

T(0): 200F

200 = Ce^(-k*0)+68

C = 132

Step 3:

T(10) = 132e^(-10k)+68

145-68 = 132e^(-10k)

77/132 = e^(-10k)

Taking natural log

k = 0.0538

Step 4:

T(15) = 132e^(-0.0538*15)+68

=126.872F

Answer = Temperature after 15 min: 126.872F

Aug 30th, 2015

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Aug 30th, 2015
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Aug 30th, 2015
Oct 21st, 2017
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