Respuesta :
Answer : The heat of this reaction will be, 2.81 kJ
Explanation :
First we have to calculate the mass of water.
[tex]\text{Mass of water}=\text{Density of water}\times \text{Volume of water}[/tex]
Given:
Density of water = 1.0 g/mL
Volume of water = 100.0 mL + 100.0 mL = 200.0 mL
[tex]\text{Mass of water}=1.0g/mL\times 200.0mL[/tex]
[tex]\text{Mass of water}=200.0g[/tex]
Now we have to calculate the heat of this reaction.
Formula used :
[tex]q=m\times c\times (T_{final}-T_{initial})[/tex]
where,
q = heat = ?
m = mass of water = 200.0 g
c = specific heat of water = [tex]4.18J/g^oC[/tex]
[tex]T_{final}[/tex] = final temperature = [tex]23.65^oC[/tex]
[tex]T_{initial}[/tex] = initial temperature = [tex]20.29^oC[/tex]
Now put all the given values in the above formula, we get:
[tex]q=200.0g\times 4.18J/g^oC\times (23.65-20.29)^oC[/tex]
[tex]q=2808.96J=2.81kJ[/tex]
Thus, the heat of this reaction will be, 2.81 kJ
