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In a neutralization reaction, is heat given out to the system, thereby raising its temperature? Suppose we put a thermometer in the acid-base mixture and there is an increase in the thermometer reading. Is this increase simply because part of the reaction's heat given out is supplied to this thermometer, or because the water solvent is actually warmer? In other words, if we measure the temperature of the system before and after the reaction, will there have been an overall decrease in the temperature, or increase (and please explain why)?

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In thermodynamics, we like to divide any situation into two parts, the system and the surroundings. In general the system is the part of the universe that is being studied while the surroundings is everything else that interacts with the system. Between the boundary of the system and the surroundings, energy and heat can be transferred (in an open system).

So in an neutralisation reaction, the system is the actual acid and base while the surroundings is the solvent (the water). Since it is an exothermic reaction, this means that energy is released from the system to the surroundings. Since the water gains more kinetic energy from the system as the reaction occurs, its temperature will increase. Also this means that the system loses energy and therefore become colder. So to answer your question, the solvent gets hotter while the system gets colder.

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  • $\begingroup$ so then if we put a thermometer in it, we'll be measuring the temperature of the water? $\endgroup$ – Airdish Nov 29 '15 at 5:01
  • $\begingroup$ Technically you will be measuring the temperature of the system and the surroundings combined, however since the solvent is in such excess, it is basically measuring the temperature of the water. $\endgroup$ – Nanoputian Nov 29 '15 at 5:55

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