# 4th Round of 29th year

### 1. ... Kofola's(2 points)

Let's have a Kofola (a Czech soft drink) with the energetic value $Q$_{$k$} = 1360 kJ ⁄ kg and temperature $t$_{$k$} = 24 ° C and another Kofola, this time sugar-free, with the energetic value $Q$_{$free$} = 14.4 kJ ⁄ kg and temperature $t$_{$free$} = 4 ° C. If we assume other behaviour and constants are very similar to water, what temperature would have a mixture of these two for which the total energy gain would be none.

### 2. ... Brain in a microwave(2 points)

How far from a base transceiver station (BTS) do a person have to be, for the emission to be fully comparable with that of the mobile phone just next to somebody's head. Expect the BTS to broadcast uniformly into a half-space with the emission power 400 W. The emission power of a mobile phone is 1 W.

### 3. ... Save the woods(3 points)

We have a toilet paper roll with the diameter $R$ = 8 cm with an inside hollow tube of diameter $r$ = 2 cm. Every layer of the paper has the thickness $d$ = 200 µm and the layers lies perfectly on top of each other. By how many does the number of pieces of the paper differ had we used a piece of the length $l$_{1} = 9 cm instead of $l$_{2} = 13 cm? A part of the solution has to be an estimate of the approximation error (if you use one).
*Bonus*: Calculate the precise length of the spiral the toilet paper makes.

### 4. ... Bubbles reunited!(4 points)

What is the smallest number of equally sized soap bubbles with the diameter $r$, that would make a single bubble with the diameter at least 3$r$? Expect the air in the bubbles to have a constant temperature.

### 5. ... Slide(5 points)

There are two identical blocks with the mass $m$ and one of the sides of lenght $l$on a horizontal plane. The distance between the closest two faces is 2$x$_{0}. Suddenly we start pouring water between them with the volume flow $Q$. At two sides of the blocks there is a barrier keeping the water in the place between the two blocks. The coefficient of static friction between the block and the plane is $f$_{0} and the of the kinetic friction is $f$. There is no friction between the barriers and the blocks. What is the condition for $f$_{0} that would keep the blocks in place? In the case of sufficiently small $f$_{0}, determine the acceleration of blocks as a function of position and also the distance, where the blocks eventually stop moving. Consider all the movement of the water reasonably slow, for any eddies to appear, for any heating of the water solely from its movement to take place or for any significant kinetic energy possesion. For the same reason of very slow $Q$, we can approximate there is no contribution of adding any water past the point where the blocks started moving. Bonus: Find a condition for turning the block over.

### P. ... Diet tower(5 points)

How tall could be a tower built from aluminium cans of diet soft drink?

### E. ... Break it down(8 points)

Measure the tensile strenght of office paper. Use a common office paper with the density 80 g·m^{−2}.

### S. ... serial(6 points)

* From the inequality

$$\Delta S_{tot} \ge 0 }$$

and given the equation from the text of the serial

$$\Delta S_{tot} = \frac{-Q}{T_H} \frac{Q-W}{T_C}$$

express $W$ and derive this way the inequality for work

$$W\le Q\left( 1 - \frac {T_C}{T_H} \right).$$

- Calculate the efficiency of the Carnot cycle without the use of entropy.

*Hint:* Write out 4 equations connecting 4 vertices of the Carnot cycle

$$p_1 V_1 = p_2 V_2 $$

$$p_2 V_2^{\kappa} = p_3V_3^{\kappa}$$

$$p_3V_3 = p_4V_4$$

$$p_4V_4^{\kappa} = p_1V_1^{\kappa}$$

and multiply all of them together. By modifying this equation you should be able to get

$$\frac {V_2}{V_1} = \frac {V_3}{V_4}.$$

Next step is using the equation for the work done in an isothermal process: when going from the volume $V$_{$A$} to the volume $V$_{$B$}, the work done on a gas is

$$nRT\,\mathrm{ln}\left(\frac{V_A}{V_B}\right).$$

Now the last thing we need to realize is that the work in an isothermal process is equal to the heat (with the correct sign) a calculate the work done by the gas (there is no contribution from the adiabatic processes) and the heat taken away.

$ For the correct solution, you only need to fill in the details.$- In the last problem you worked with $pV$ and $Tp$ diagram. Do the same with $TS$ diagram, i. e. sketch there the isothermal, isobaric, isochoric and adiabatic process. In addition sketch the path for the Carnot cycle including the direction and labeling of the individual processes.

- Sometimes it is important to check if we give or receive heat. Because sometimes this fact can change during the process. One of the examples is the process

$$p=p_0\mathrm{e}^{-\frac{V}{V_0}},$$

where $p$_{0} and $V$_{0} are constants. Show for which values of $V$ (during the expansion) the heat is going into the gas and for which out of it.