According to zeroth law of thermodynamics, when two objects are kept in contact, heat (energy) is transferred from one to the other until they reach the same temperature (are in thermal equilibrium). When the objects are at the same temperature there is no heat transfer.
So, at equilibrium, =
,
+
q=m×c×T, where q = heat energy, m = mass of a substance, c = specific heat (units J/kg∙K), T is temperature
=(15X13X4.19)+(148X88.3X4.19)
= 81.37 ° C
Answer : The enthalpy of neutralization is, 56.012 kJ/mole
Explanation :
First we have to calculate the moles of HCl and NaOH.
The balanced chemical reaction will be,
From the balanced reaction we conclude that,
As, 1 mole of HCl neutralizes by 1 mole of NaOH
So, 0.08 mole of HCl neutralizes by 0.08 mole of NaOH
Thus, the number of neutralized moles = 0.08 mole
Now we have to calculate the mass of water.
As we know that the density of water is 1 g/ml. So, the mass of water will be:
The volume of water =
Now we have to calculate the heat absorbed during the reaction.
where,
q = heat absorbed = ?
= specific heat of water =
m = mass of water = 400 g
= final temperature of water =
= initial temperature of metal =
Now put all the given values in the above formula, we get:
Thus, the heat released during the neutralization = -4480.96 J
Now we have to calculate the enthalpy of neutralization.
where,
= enthalpy of neutralization = ?
q = heat released = -4480.96 J
n = number of moles used in neutralization = 0.08 mole
The negative sign indicate the heat released during the reaction.
Therefore, the enthalpy of neutralization is, 56.012 kJ/mole