The energy [tex]6.311 \times 10^{7} \mathrm{J}[/tex] is transferred to the surroundings from water in order to freeze completely.
Explanation:
Water will transfer to surrounding will come from cooling energy from 22.3°C to 0°C and then freezing energy is
[tex]\mathrm{E}_{\mathrm{t}}=\mathrm{E}_{\text {cooling }}+\mathrm{E}_{\text {freezing }}[/tex]
[tex]\mathrm{E}_{\mathrm{t}}=\mathrm{M}\left(\mathrm{C}_{\mathrm{w}} \Delta \mathrm{t}+\mathrm{L}_{\mathrm{f}}\right)[/tex]
We know that,
\mathrm{C}_{\mathrm{W}}=4190 \mathrm{J} / \mathrm{kgk}
[tex]\mathrm{L}_{\mathrm{f}}=333 \times 10^{3} \mathrm{J} / \mathrm{kg}[/tex]
As per given question,
M = 148 kg
[tex]\Delta \mathrm{t}=22.3^{\circ} \mathrm{C}[/tex]
Substitute the values in the above formula,
[tex]\mathrm{E}_{\mathrm{t}}=148\left(4190 \times 22.3+333 \times 10^{3}\right)[/tex]
[tex]E_{t}=148\left(93437+333 \times 10^{3}\right)[/tex]
[tex]E_{t}=148 \times 426437[/tex]
[tex]\mathrm{E}_{\mathrm{t}}=6.311 \times 10^{7} \mathrm{J}[/tex]
The energy [tex]6.311 \times 10^{7} \mathrm{J}[/tex] is transferred to the surroundings from water in order to freeze completely.
