Notes

 

Practice Problems

A cup of tea at \(180^{\circ}\) F, and the room temperature is \(70^{\circ}\) F. After 5 minutes, the temperature of the tea is \(152^{\circ}\) F.

\(\textbf{1)}\) Solve for k in Newtons law of cooling formula.

Show Work and Answer

\(152= \left( 180-70 \right) e^{k(5)}+70\)
\(152= \left( 110 \right) e^{k(5)}+70\)
\(82= \left( 110 \right) e^{k(5)}\)
\(\frac{82}{110}= e^{k(5)}\)
\(\ln{\frac{82}{110}}=5k\)
\(\displaystyle\frac{\ln{\frac{82}{110}}}{5}=k\)
\(k \approx -0.0587522237\)

\(\textbf{2)}\) What is the temperature of the tea after 10 minutes?

Show Answer

\(y_{10}= \left( 180-70 \right) e^{-0.0587522237(10)}+70\)
\(y_{10}= 131.14^{\circ}\)

\(\textbf{3)}\) What is the temperature of the tea after 20 minutes?

Show Answer

\(y_{20}= \left( 180-70 \right) e^{-0.0587522237(20)}+70\)
\(y_{20}= 103.97^{\circ}\)

 

See Related Pages\(\)

\(\bullet\text{ Calculus Homepage}\)
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\(\bullet\text{ Expanding Logarithms}\)
\(\,\,\,\,\,\,\,\,2\log_{b}(x)+\log_{b}(z)-5\log_{b}(y)…\)

\(\bullet\text{ Decibel Problems}\)
\(\,\,\,\,\,\,\,\,N_{dB}=10\log \left(\frac{P}{10^{-12}}\right)…\)

\(\bullet\text{ Earthquake Problems}\)
\(\,\,\,\,\,\,\,\,M=\log\frac{I}{10^{-4}}…\)

\(\bullet\text{ Domain and Range Logarithmic Functions}\)
\(\,\,\,\,\,\,\,\,f(x)=log(x) \rightarrow \text{Domain:} x\gt0… \)

\(\bullet\text{ Graphing Logarithmic Functions}\)
\(\,\,\,\,\,\,\,\,f(x)=log_{2}(x)\)

\(\bullet\text{ Solving Logarithmic Equations}\)
\(\,\,\,\,\,\,\,\,\log_{2}(5x)=\log_{2}(2x+12)…\)

\(\bullet\text{ Inverse of Logarithmic Functions}\)
\(\,\,\,\,\,\,\,\,f(x)=log_{2}(x) \rightarrow f^{-1}(x)=2^x\)

 

In Summary

Newton’s Law of Cooling is a scientific principle that describes the rate at which an object cools down. The rate of heat loss of an object is directly proportional to the temperature difference between the object and its surroundings.