Which Cup Will Be Hottest After 15 Minutes?

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SUMMARY

The discussion centers on the thermal dynamics of two cups of coffee prepared differently and their temperatures after 15 minutes. Cup 1, which has milk added initially, will be slightly hotter than Cup 2, which has milk added later. This conclusion is based on the principles of heat loss, where the higher initial temperature of Cup 2 results in greater heat loss over time. The mathematical analysis confirms that mixing at the beginning leads to a higher final temperature due to slower cooling rates.

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this was in a quiz i did ages ago and i never found out the answer.

someone makes 2 cups of coffee. Cup 1 has coffee then water then milk added and is left to stand for 15mins. Cup 2 has coffee then water added then is left to stand for 15 mins, then milk is added. which one would be hottest?

i think it would be cup 2 but I am not sure and i can't really explain it properly either.

thank to anyone who can help, I've been trying to work it out for ages
 
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i think it would be cup 2
Wrong guess.

The greater is the temperature then the greater is the rate of heat loss. This means that "cup 2" which is not initially cooled by the addition of milk will be at a higher average temperature for the duration of the 15 minute period. It will therefore have lost more heat during that time and after the milk is eventually added it will surely be cooler than cup 1.
 
let's see what is going on here
Let's [tex]T_1[/tex] to be the hot water temperature, [tex]T_2[/tex] to be
milk (room) temperature, [tex]\kappa[/tex] to be the thermal conductivity, (which we assume to be constant) [tex]\tau[/tex] =15 min, [tex]C_1[/tex] and [tex]C_2[/tex] to be the heat capacity of water and milk.

1) if we mix the hot water with milk at the beginning

then the mixture temperature is
[tex]T_m =[tex](C_1T_1 +C_2 T_2)/(C_1+C_2[/tex]<br /> <br /> The temperature after 15 minutes will be<br /> <br /> <br /> [tex]T_{f1} =(T_m-T_2)exp^{-\kappa \tau/C_1+C_2}+T_2=(C_1/C_1+C_2)(T_1-T_2)exp^{-\kappa \tau/C_1+C_2)}+T_2[/tex]<br /> <br /> <br /> 2)if we mix afterwards<br /> the water temperature after 15 minutes will be<br /> [tex]T_3=(T_1-T_2)exp^{-\kappa \tau/C_1}+T_2[/tex]<br /> <br /> The temperature after mixing will be<br /> <br /> [tex]T_{f2}=(C_1T_3 +C_2 T_2)/(C_1+C_2)[/tex]=<br /> =[tex](C_1/C_1+C_2)(T_1-T_2)exp^{-\kappa \tau/C_1}+T_2[/tex]<br /> <br /> <br /> Now because [tex]exp^{-\kappa \tau/C_1[/tex]<[tex]exp^{-\kappa\tau/C_1+C_2[/tex], then [tex]T_{f1}>T_{f2}[/tex]<br /> <br /> the answer: the if we mix at the beginning, the cup will be slightly hotter, because it will take mor time for cup to cool. <br /> <br /> We assumed that the thermal conductivity does not depend on the volueme of the liquid. Actually, for the metal cup the thermal loss will be higher for the cup with milk, and both cups could have the same temperature at the end. It is also possible for cup 2 to be hotter as well. As to validity of summation of the specific heat of water and milk, it is justified because milk is actually a mixture itself, and no chemical reaction occurs if we mix milk with water.[/tex]
 

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