We fill a flask with one kilogram of water and stir it with an Ultrapure agitato
ID: 1050788 • Letter: W
Question
We fill a flask with one kilogram of water and stir it with an Ultrapure agitator. This is a piece of laboratory equipment for vigorous stirring. The stirring power or work flow is 1 kW and we stir for one minute: How much does the internal energy of the water increase?; How much does the temperature increase? The c_v of water is 4185 J Kg^-1 K^-1. What are the consequences of this for heat sensitive materials during stirring (example: enzymes)?; Instead of stirring, we heat our kilogram of water. The heat flow is again 1 kW and the heating time one minute. What are now the changes in internal energy and temperature? Sol: 60kJ/kg; delta T = 14.3 K; similar.Explanation / Answer
a) The power of the stirring apparatus is 1 kW = 1 kJ/sec and we stir for 1 minute = 60 secs. Therefore, the work done by the stirrer is W = (Power of the stirrer)*(time of stirring) = (1 kJ/1 sec)*(60 sec) = 60 kJ.
This work is converted into internal energy of the water; therefore the increase in internal energy of the water = 60 kJ. Since we have 1 kg of water, the change in internal energy is given as 60 kJ/kg (ans).
b) Let the change in temperature of the water be T K. Therefore, using the principle of thermochemistry, the change in internal energy = heat gained by the water = (mass of water)*(sp.heat capacity of water)*(change in temperature).
Therefore,
60 kJ = (1 kg)*(4185 J kg-1 K-1)*T
=====> T = (60*1000 J)/(4185 J K-1) = 14.336 K 14.3 K (ans).
The change in temperature is significant. For heat sensitive materials like enzymes, an increase in temperature by 14.3 K can lead to undesired reactions including decomposition or denaturation of the enzyme.
c) The effect will be similar because now 1 kW = 1 kJ/sec heat is flowing into the water and the heating time is 1 minute = 60 sec. The heat gained by the water will be 60 kJ and hence the change in internal energy and the rise in temperature will be the same.