21H = 2αCosθ/RÏg
mgH/2 = VαCosθ/R
V = H*Î R2 = (2αCosθ/RÏg)* Î R2 = 2αCosθΠR/Ïg
ther4 gain in PE = (2αCosθΠR/Ïg)αCosθ/R
ther4 Heat evolved = 2α2Cos2θΠ/Ïg
If a capillary tube is dipped in a liquid of density \rho with surface tension \alpha, what is the amount of heat evolved? Take the contact angle to be \theta.
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9 Answers
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66@tapan
sorry to reply late...
What you are saying is that the heat evolved is same as gravitational potential energy increment of the liquid column. Is it consistent with the energy conservation? There is some heat coming out of the system and the potential energy is also rising. Who is supplying the total energy required?
21OH !!!
one possible energy supplier cud B SURFACE ENERGY = T∂A ??
21ummmm..............
Change in area will be the contact surface created due to the rise
so dA = 2Î RH
wer H = H = 2αCosθ/RÏg
so dW = 4ΠαCosθ/Ïg
dU = 2α2Cos2θΠ/Ïg
dQ = dU + dW
dQ = HEAT LIBERATED!!
66Tapan you are again applying the energy conservation incorrectly. In the above, dQ is the heat that should be given to the system and not the heat evolved.
You see, when the water rises, the force of surface tension does some work. This work, is simply
W=(2\pi r\alpha\cos\theta) h
The quantity inside the brackets is the vertical component of the force due to surface tension. Using the fact thath=\dfrac{2\alpha \cos\theta}{r\rho g}, we get the work done by the force of surface tension as
W=\dfrac{4\pi \alpha^2\cos^2\theta}{\rho g}
On the other hand, the increment in the potential energy of the water column is
U=(\rho\pi r^2h)g\dfrac{h}{2}=\dfrac{2\pi \alpha^2\cos^2\theta}{\rho g}
We see that W=2U, so that only half of the work done by the surface tension has been utilized to raise the gravitational potential energy. The remaining half, therefore, must be lost as heat. And so the heat evolved is
Q=\dfrac{2\pi \alpha^2\cos^2\theta}{\rho g}