Can Holding a Teacup Make Tea Cool Faster?

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SUMMARY

Holding a teacup with both hands significantly accelerates the cooling process of tea compared to letting it sit. The heat transfer from the tea to the hands is more efficient due to the higher density of the human body compared to air, which allows for a greater heat transfer constant in the differential equation governing temperature change. Experimental calculations using the equations T1 and T2 demonstrate that the time taken for the tea to cool is shorter when held, with specific values indicating t_h (holding) is approximately 1.14 compared to t_c (not holding) at approximately 1.29. Therefore, holding the cup does indeed facilitate faster cooling of the tea.

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  • Understanding of basic thermodynamics and heat transfer principles.
  • Familiarity with differential equations and their application in temperature change.
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  • Basic grasp of the concepts of thermal conductivity and heat transfer coefficients.
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  • Explore the principles of heat transfer in solids versus gases.
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Which cup will cool faster


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Harut82
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We were having tea with the family and my brother and I started wondering if holding the cup will help in cooling the tea.

Lets say you have two identical cups filled with the same amount of the same temperature tea.

You let one just sit and the other you grab with both hands . . . which tea will cool quicker?

When you hold a hot teacup you feel heat transfering to your hand. I argued that holding the cup with both hands will help it cool faster.

Temperature of hand > temperature of surrounding air near cup

therefore tea cools faster


am I wrong?
 
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Depends on the temperature of the air and of your hand.
 
One way to know for sure. Find a thermometer and do the experiment.
 
Alex48674 said:
Depends on the temperature of the air and of your hand.

Air is room temperature. Let's say 72 degrees.
Body temperature 98.6 degrees.
 
I'd suspect that the constant in the differential equation, caused by your hands, would lower the temperature faster. If you look at the differential equation,
T'=k(T_e-T)
it would seem that a mere doubling in the constant (which is not unreasonable in going from air to your hands) would have a much more profound affect on the time derivative of temperature than the lower temperature of the air, assuming T_o>130. Just a thought, test it to see for sure.
 
Would I be right in saying a solid like your hand would be much more capable of transporting heat than air due to its higher density. Is this why you would increase the constant in the differential?

Of course the cup wouldn't attain the same equilibrium temperature as your hadns are warmer than the air. Also the transfer of heat to air from your hands would have to be taken into account and its conduction through your arms. I suspect if you have a very big cup of tea that you are not able to grasp properly then it wouldn't make much difference :)

But for a normal cup I would say it would cool faster with both hands.

Alex
 
twinsen said:
Would I be right in saying a solid like your hand would be much more capable of transporting heat than air due to its higher density. Is this why you would increase the constant in the differential?
Yea, that was my reasoning and it holds for any cup. Substituting a small portion of a huge cup with your hands would still help lower the temperature more quickly. We can be more quantitative, T1 is the temp of the cup without the hands, T2 is with the hands. The solution the the differential equations are:
T_1=T_c+(T_o-T_c)e^{-k_ct}
where:
-T_c is the temp of the cup
-T_o is the initial temp of the tea
-k_c is the constant between the cup and the tea.
and
T_2=T_h+(T_o-T_h)e^{-k_ht}
where everything is the same as above but the h subscript means hands.

We need to take into account the value of the temperature that you find fitting for the tea, which we call T_r. Set both equations equal to that and solve for t. The results are:
t_c=\frac{1}{k_c}\log\frac{T_o-T_c}{T_r-T_c}
t_h=\frac{1}{k_h}\log\frac{T_o-T_h}{T_r-T_h}.
If we substitute some reasonable values in (i use fahrenheit), T_r=110, T_o=210, T_h=98.6, T_c=72, and k_c=1,k_h=2 (it's not unreasonable for the hand constant to be twice the cup constant, btw 1 and 2 are WAY too high for the constants, but the ratio 1/2 is all that matters) we find,
t_c\approx1.29
and
t_h\approx1.14.
So the hands cool it faster in this case.
 
Trick is, final temperature with cup hold in hands is higher, so in fact final answer depends on the way you define the speed - at higher temps it will be faster but at lower temps it will be slower.

And let's not forget that your hands are not just solid, you also have an internal heating/cooling system so the answer of your body (heat transfer speed) is non-linear.
 

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