Answer:
[tex]Q=0.41kcal[/tex]
Explanation:
Hello.
In this case, since the energy involved during this process is associated to the freezing enthalpy of water at the freezing point of water (0.0 °C) and is computed via:
[tex]Q=m\Delta _{freezing}H[/tex]
It is known that the freezing enthalpy of water is -333.3 J/g, therefore, the involved heat is:
[tex]Q=5.2g*-333.33\frac{J}{g}\\ \\Q=-1733.33J[/tex]
Clearly, it is negative since freezing requires the removal of energy, which means it is negative based on the first law of thermodynamics. Finally, the energy in kcal turns out:
[tex]Q=1733.33J*\frac{1kcal}{4184kJ}\\ \\Q=0.41kcal[/tex]
Regards.
