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To find the enthalpy change of combustion for an alcohol, you use a calorimeter filled with water and heat it by burning an alcohol.

I was told that usually experimental values for the enthalpy change of combustion are less exothermic than data book values, because of many reasons (eg- heat loss, incomplete combustion etc.) One of the reasons why the experimental value may be less exothermic is because water has evaporated from the calorimeter.

I don't understand how water evaporating from the calorimeter would make the enthalpy change less exothermic. If the water evaporates, then the mass of water left in the calorimeter is lower, so it would heat up faster, so wouldn't this give a more exothermic result?

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    $\begingroup$ Heat of vaporization is quite high, so if in the local area of the combustion, you vaporize water, that heat won't be distributed to the rest of water to warm it up. $\endgroup$ – Zhe May 18 at 16:42
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    $\begingroup$ When water evaporates the heat to vaporize the water comes from the water itself, so the water cools. // Apologies to the "young" chemists, but I still think in calories. It takes 540 cal/gram to evaporate water, but 1 cal/g to change water temperature by $\pu{1 ^\circ C}$. $\endgroup$ – MaxW May 18 at 19:04
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First, we have some things to consider and they are:

  1. the specific heat of water;
  2. the latent heat of vaporization of water;
  3. how the terms mentioned above are compared to each other.

The specific heat of water is $\pu{4200 J kg-1 K-1}$ and it means that it takes $\pu{4200 kJ}$ amount of heat to increase the temperature of $\pu{1 kg}$ water by $\pu{1 K}.$ On the other hand, the amount of heat needed to turn $\pu{1 kg}$ water at $\pu{373 K}$ into $\pu{1 kg}$ water vapor is $\pu{2268000 J kg-1},$ i.e. the latent heat of vaporization of water is $\pu{2268000 J kg-1}.$

Now, if we consider two pieces of data above, it's clearly visible that it takes much more heat to vaporize water than it takes to increase the temperature of the water (given the same mass). As it's this heat which absorbs heat from the reaction materials (mainly products), the high latent heat of vaporization makes the reaction much less exothermic than increasing the temperature of the water (since it requires more heat).

So, reducing the exothermic characteristic of the reaction depends more on the rate of vaporization than it does on the increase in temperature of the water. As a result of this, despite the fact that the amount of water available to absorb heat via increasing its temperature decreases as more and more water is vaporized, the amount of heat absorbed by vaporization is more than enough to compensate for this. Thus, the reaction becomes less exothermic than expected with the vaporization of water.

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