Question

You place a cup of 205oF coffee on a table in a room that is 72oF, and 10 minutes later, it is 195oF. Approximately how long will it be before the coffee is 180oF? Use Newton's law of cooling: \[T(t)=T _{A}+(T _{o}-T _{A})e ^{-kt}\]
A. 15 minutes
B. 25 minutes
C. 1 hour
D. 45 minutes

Answer 1

Answer:

The coffee temperature will be 180 ° F in approximately 25 minutes.

Explanation:

Newton's cooling law states that, when the temperature difference between a body and its environment is not too large, the heat transferred per unit of time to the body or from the body by conduction, convection and radiation, is approximately proportional to the difference in temperatures between the body and said external medium, as long as the latter keeps its temperature constant during the cooling process.

This law is manifested by the following equation

$T(t)=TA+(T0-TA)*exp^{-k*t}$

where k is a constant of proportionality known as the cooling parameter, TA is the ambient temperature, which is always assumed constant, and T0 is the temperature at the instant zero, or initial temperature.

First, it is convenient to determine the value of the constant k. For that, knowing that a 205 ° F coffee cup is placed on a table in a 72 ° F room, and 10 minutes later, the temperature is 195 ° F. These data, then, are:

• T0=205°F
• TA=72°F
• t=10minutes
• T(t)=195°F

By replacing, you can get the value of the constant k:

$195=72+(205-72)*exp^{-k*10}$

$123=133*exp^{-k*10}$

$\frac{123}{133} =exp^{-k*10}$

$ln(\frac{123}{133} )=-10*k$

$k=-\frac{1}{10} *ln(\frac{123}{133} )$

With the value of k, the expression of Newton's cooling law is:

$T(t)=72+(205-72)*exp^{\frac{1}{10} *ln(\frac{123}{133} )*t}$

$T(t)=72+133*exp^{\frac{1}{10} *ln(\frac{123}{133} )*t}$

You are looking for the amount of time it takes for coffee to cool down to a temperature of  180  Fahrenheit degrees. This value is T (t). Then the equation is:

$180=72+133*exp^{\frac{1}{10} *ln(\frac{123}{133})*t }$

$108=133*exp^{\frac{1}{10} *ln(\frac{123}{133})*t }$

$\frac{108}{133} =exp^{\frac{1}{10} *ln(\frac{123}{133})*t }$

$ln(\frac{108}{133} )=\frac{1}{10} *ln(\frac{123}{133})*t$

$\frac{10*ln(\frac{108}{133} )}{ln(\frac{123}{133}) } =t$

Then the value of time t is:

t≈26.64 min

Taking into account the options given, the coffee temperature will be 180 ° F in approximately 25 minutes.

Answer 2

Temp first number is 205−72=133
second number is 195−72=123

This takes 10 minutes to occur, so you can model this as:133(123133) t(10)