Respuesta :
Cool liquid from 314 K to 273 K, freeze liquid at 273 K, and cool solid to 263 K.
The amount of heat released in the given situation would be as follows:
[tex]-66.2 KJ[/tex]
Given that,
Weight of [tex]H2O[/tex] [tex]= 27g[/tex]
The Temperature of the Liquid [tex]= 314 k[/tex]
The Temperature of the Solid [tex]= 263 K[/tex]
Melting point of [tex]H2O = 273 K[/tex]
To find,
The amount of heat released = [tex][Q[/tex] × [tex]C_{p}l[/tex] ×[tex](T_{final} - T_{Initial} ) +[/tex] Δ [tex]Hfusion + [m[/tex] × [tex]C_{p}s[/tex] × [tex]T_{final} - T_{Initial} ][/tex]
By putting the values in the above formula,
∵ Heat released = [tex]-66.2 KJ[/tex]
Learn more about "Melting Point" here:
brainly.com/question/5753603
