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Calculate the amount of heat released when 27.0 g H2O is cooled from a liquid at 314 K to a solid at 263 K. The melting point of H2O is 273 K. Other useful information about water is listed below.


Cp,liquid = 4.184 J/(g•K)


Cp,solid = 2.093 J/(g•K)


ΔHfusion = 40.7 kJ/mol

Verified Answer :

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:

\(-66.2 KJ\)

Given that,

Weight of \(H2O\) \(= 27g\)

The Temperature of the Liquid \(= 314 k\)

The Temperature of the Solid \(= 263 K\)

Melting point of \(H2O = 273 K\)

To find,

The amount of heat released = \([Q\) × \(C_{p}l\) ×\((T_{final} - T_{Initial} ) +\) Δ \(Hfusion + [m\) × \(C_{p}s\) × \(T_{final} - T_{Initial} ]\)

By putting the values in the above formula,

∵ Heat released = \(-66.2 KJ\)

Learn more about "Melting Point" here:

brainly.com/question/5753603

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