Respuesta :
Answer : The amount of heat released, -35057 J
Solution :
Process involved in the calculation of heat released :
[tex](1):H_2O(g)(100^oC)\rightarrow H_2O(l)(100^oC)\\\\(2):H_2O(l)(100^oC)\rightarrow H_2O(l)(25^oC)[/tex]
Now we have to calculate the amount of heat released.
[tex]Q=\Delta H_{condensation}+[m\times c_{p,l}\times (T_{final}-T_{initial})][/tex]
where,
Q = amount of heat released = ?
m = mass of water = 18 g
[tex]c_{p,l}[/tex] = specific heat of liquid water = [tex]4.18J/g^oC[/tex]
[tex]\Delta H_{condensation}[/tex] = enthalpy change for condensation = -40700 J/mole
Now put all the given values in the above expression, we get
[tex]Q=-40700J+[18g\times 4.18J/g^oC\times (100-25)^oC][/tex]
[tex]Q=-35057J[/tex]
Therefore, the amount of heat released, -35057 J
Answer: 46418.4 Joules
Explanation:
The conversions involved in this process are :
[tex]H_2O(g)(100^0C)\rightarrow H_2O(l)(100^0C)[/tex]
[tex]H_2O(l)(100^0C)\rightarrow H_2O(l)(25^0C)[/tex]
Now we have to calculate the enthalpy change.
[tex]Q=[[m\times \Delta H_{condensation}]+[m\times c_{p,l}\times (T_{final}-T_{initial})][/tex]
where,
Q = amount of heat released = ?
m = mass of water = 18 g
[tex]c_{p,l}[/tex] = specific heat of liquid water = [tex]4.18J/g^oC[/tex]
[tex]\Delta H[/tex] = enthalpy change = ?
m = mass of water = 25 g
[tex]\Delta H_{condensation}[/tex] = enthalpy change for condensation = 2265 J/g
Now put all the given values in the above expression, we get
[tex]Q=[(18g\times 2265J/g)+(18g\times 4.18J/g^oC\times (100-25))^oC][/tex]
[tex]Q=40770+5648.4=46418.4J[/tex]
Therefore, the amount of heat released is 46418.4 J.
