This problem is taken from the second IphO Budapest, Hungary, Theoretical problems, in 1968
Problem 2
There are 300 cm^{3} toluene of 0°C temperature in a glass and 110 cm^{3} toluene of 100°C temperature in another glass. (The sum of the volumes is 410 cm°C .) Find the final volume after the two liquids are mixed. The coefficient of volume expansion of toluene β = 0.001(°C)^{−1} . Neglect the loss of heat.
Solution
If the volume at temperature t_{1} is V_{1}, then the volume at temperature 0°C is V_{10} = V_{1}/(1+βt_{1}). In the same way if the volume at t_{2} temperature is V_{2}, at 0°C we have V_{20} = V_{2}/(1+βt_{2}). Furthermore if the density of the liquid at 0°C is d, then the masses are m_{1} = V_{10} d and m_{2} = V_{20} d, respectively. After mixing the liquids the temperature is
The volumes at this temperature are V_{10} (1+βt_{2}) and V_{20} (1+βt_{2}). The sum of the volumes after mixing:
The sum of the volumes is constant. In our case it is 410 cm^{3}. The result is valid for any number of quantities of toluene, as the mixing can be done successively adding always one more glass of liquid to the mixture.
Sources/Literatures :
-Waldemar Gorzkowski, Institute of Physics, Polish Academy of Sciences, Warsaw, Poland
-Péter Vankó, Institute of Physics, Budapest University of Technical Engineering, Budapest, Hungary, Problems of the 2nd and 9th International Physics Olympiads (Budapest, Hungary, 1968 and 1976)
-The Olympic home page www.jyu.fi/ipho