Solving Newton's Cooling Law Homework

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SUMMARY

The discussion focuses on solving a homework problem related to Newton's Law of Cooling, specifically determining the cooling rate, k, for a cup of hot chocolate cooling from 86°C to 65°C in 15 minutes. The solution involves applying the formula T - Ts = (T0 - Ts)e^(kt), where Ts is the surrounding temperature. The calculated value of k is approximately -0.02601, confirming the accuracy of the solution. Participants affirm the correctness of the approach and the use of logarithmic functions in the calculations.

PREREQUISITES
  • Understanding of Newton's Law of Cooling
  • Familiarity with natural logarithms and their properties
  • Basic knowledge of exponential functions
  • Ability to solve algebraic equations
NEXT STEPS
  • Study the derivation of Newton's Law of Cooling
  • Learn about differential equations related to cooling processes
  • Explore applications of exponential decay in real-world scenarios
  • Investigate the impact of varying ambient temperatures on cooling rates
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Students studying physics or mathematics, educators teaching thermodynamics, and anyone interested in the practical applications of cooling laws in science.

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Homework Statement


"A cup of hot chocolate, in a room temperature of 21*C, cools accordiing to Newton's law of cooling. Determine the rate of cooling, k, of the hot chocolate if it cools from 86*C to 65*C in 15 minutes"


Homework Equations


Newton's cooling law, ln function
T-Ts=(T. - Ts)e^(kt)


The Attempt at a Solution



T-Ts=(T. - Ts)e^(kt)
65-21=(86-21)e^(15k)
44=65e^(15k)
44/65=e^(15k)
ln(44/65)=15k
(ln(44/65))/15=k
k=-0.02601...
Is this right?
 
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venger said:

Homework Statement


"A cup of hot chocolate, in a room temperature of 21*C, cools accordiing to Newton's law of cooling. Determine the rate of cooling, k, of the hot chocolate if it cools from 86*C to 65*C in 15 minutes"

Homework Equations


Newton's cooling law, ln function
T-Ts=(T. - Ts)e^(kt)

The Attempt at a Solution



T-Ts=(T. - Ts)e^(kt)
65-21=(86-21)e^(15k)
44=65e^(15k)
44/65=e^(15k)
ln(44/65)=15k
(ln(44/65))/15=k
k=-0.02601...
Is this right?
were you taught this formula or the differential equation?

it's right btw
 

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