Answer: The amount of heat released when 233 grams of water at room temperature (25C) is frozen into ice at -15C is 17253.65 J.
Explanation:
As water converts into ice at [tex]0^{o}C[/tex]. So, the change in temperature is taking place from [tex]25^{o}C[/tex] to [tex]0^{o}C[/tex] and then from [tex]0^{o}C[/tex] to [tex]-15^{o}C[/tex].
So, amount of heat released by water is as follows.
[tex]q_{1} = m \times C \times (T_{2} - T_{1})\\= 233 g \times 4.18 J/g^{o}C \times (25 - 0)^{o}C\\= 24348.5 J[/tex]
m = mass of substance
C = specific heat of water = [tex]4.18 J/g^{o}C[/tex]
[tex]C'[/tex] = specific heat of ice = [tex]2.03 J/g^{o}C[/tex]
The amount of heat released by ice is as follows.
[tex]q_{2} = m \times C' \times (T'_{2} - T'_{1})\\= 233 g \times 2.03 J/g^{o}C \times (-15 - 0)^{o}C\\= -7094.85[/tex]
Now, total heat released during this change is as follows.
[tex]\Delta H = q_{1} + q_{2}\\= 24348.5 J + (-7094.85) J\\= 17253.65 J[/tex]
Thus, we can conclude that heat released when 233 grams of water at room temperature (25C) is frozen into ice at -15C is 17253.65 J.
