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?
please say step by step
Thank you for the opportunity to help you with your question!
i got 127 degrees, using the formula T=Ambient temp +(original temp -ambeint temp)e^-kt
But if you think it logically, my above calculations would be correct. Not using newton's law of cooling.
can you do it using newtons law of cooling, that is how my teacher wants it to be done
K let me try it using Newton's law of cooling.
So here is the step by step explanation for your question:
Newton's Cooling Law: T(a)=Ce^(-kt)+Ta
200 = Ce^(-k*0)+68
C = 132
T(10) = 132e^(-10k)+68
145-68 = 132e^(-10k)
77/132 = e^(-10k)
Taking natural log
k = 0.0538
T(15) = 132e^(-0.0538*15)+68
Answer = Temperature after 15 min: 126.872F
Hope I was helpful my friend, please revert if you have more questions. I would love to help you more.
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