Question

Based on the thermodynamic properties provided for water, determine the amount of energy released for 155.0 g of water to go from 39.0 °C to -36.5°C. Property Melting point Boiling point AHfus AHvap Cp (s) Value 0.0 100.0 6.01 40.67 37.1 75.3 33.6 Units oC kJ/mol kJ/mol J/mol.oc J/mol C mol oC Cp (g)

Answer 1

Answer : The amount of energy released will be, -88.39 kJ

Solution :

The process involved in this problem are :

$(1):H_2O(l)(39.0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(s)(0^oC)\\\\(3):H_2O(s)(0^oC)\rightarrow H_2O(s)(-36.5^oC)$

The expression used will be:

$\Delta H=[m\times c_{p,l}\times (T_{final}-T_{initial})]+m\times (-\Delta H_{fusion})+[m\times c_{p,l}\times (T_{final}-T_{initial})]$

where,

$\Delta H$ = heat available for the reaction = ?

m = mass of water = 155.0 g

$c_{p,s}$ = specific heat of solid water = $2.01J/g^oC$

$c_{p,l}$ = specific heat of liquid water = $4.18J/g^oC$

= enthalpy change for fusion = $6.01kJ/mole=6010J/mole=\frac{6010J/mole}{18g/mole}J/g=333.89J/g$

Molar mass of water = 18 g/mole

Now put all the given values in the above expression, we get:

$\Delta H=[155.0g\times 4.18J/g^oC\times (0-(39.0))^oC]+155.0g\times -333.89J/g+[155.0g\times 2.01J/g^oC\times (-36.5-0)^oC]$

$\Delta H=-88392.625J=-88.39kJ$

Therefore, the amount of energy released will be, -88.39 kJ