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?
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, [tex]T'=k(T_e-T)[/tex] 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 [itex]T_o>130[/itex]. 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 wouldnt 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 wouldnt make much difference :) But for a normal cup I would say it would cool faster with both hands. Alex
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: [tex]T_1=T_c+(T_o-T_c)e^{-k_ct}[/tex] where: -[itex]T_c[/itex] is the temp of the cup -[itex]T_o[/itex] is the initial temp of the tea -[itex]k_c[/itex] is the constant between the cup and the tea. and [tex]T_2=T_h+(T_o-T_h)e^{-k_ht}[/tex] 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 [itex]T_r[/itex]. Set both equations equal to that and solve for t. The results are: [tex]t_c=\frac{1}{k_c}\log\frac{T_o-T_c}{T_r-T_c}[/tex] [tex]t_h=\frac{1}{k_h}\log\frac{T_o-T_h}{T_r-T_h}[/tex]. If we substitute some reasonable values in (i use fahrenheit), [itex]T_r=110, T_o=210, T_h=98.6, T_c=72[/itex], and [itex]k_c=1,k_h=2[/itex] (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, [tex]t_c\approx1.29[/tex] and [tex]t_h\approx1.14[/tex]. 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.