Question

Constants

Heat of fusion (?Hfus) is used for calculations involving a phase change between solid and liquid, with no temperature change. For H2O, ?Hfus=6.02 kJ/mol.

Specific heat capacity (C) is used for calculations that involve a temperature change, but no phase change. For liquid water, C=4.184 J/(g??C).

Heat of vaporization (?Hvap) is used for calculations involving a phase change between liquid and gas, with no temperature change. For H2O, ?Hvap=40.7 kJ/mol.

Part B

How much heat is required to raise the temperature of 92.0g of water from its melting point to its boiling point?

Express your answer numerically in kilojoules.

Answer 1

Clearly, here no phase change is involved. We just need to increase the temperature of 92.0 g water from 0 degC (melting pt) to 100 degC (boiling pt).

So the amount of heat reqd = (final temp- initial temp) x mass of water in gram x specific heat capacity of water per gram per ?C

=(100-0) ?C x 92.0 g x 4.184 joules/ (g. degC)

=38492.8 joules

=38.4928 kilojoules

38.5 kilojoules.

However, a very related question which is often asked in this concern is,

what is the amount of heat reqd to convert y gram of ice at 0 degC to y g of steam at 100 degC?

In that case, we need to consider H_fusion & H_{vapourisation} also.

Then, total heat reqd if y=92 g is------> total H_fusionfor 92 g water+ total H_{reqd to incr temp from 0 to 100 degC}+total H_{vapourisation}

total H_fusion =5.11 x 6.02 KJ (H_fusion = 6.02 KJ/Mol; 1 KJ= 1000 J; 92 g water= 92/18 = 5.11 moles of water)

= 30.76 KJ

total H_{vapourisation} = 5.11 x 40.7 KJ =415.95 KJ

Hence total amount of heat reqd to convert 92 g of ice at 0 degC to 92 g of steam at 100 degC = (30.76 + 38.5+415.95) KJ = 485.21 KJ.

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