What is the total amount of heat released when 94.0 g water at 80.0 °C cools to form ice at −30.0 °C?

Relation between the amount of heat absorbed or released by a substance, \(q\), and its accompanying temperature change, \(\Delta T\), was introduced:

\(q = mc\Delta T\)

where \(m\) is the mass of the substance and \(c\) is its specific heat. The relation applies to matter being heated or cooled, but not undergoing a change in state. When a substance being heated or cooled reaches a temperature corresponding to one of its phase transitions, further gain or loss of heat is a result of diminishing or enhancing intermolecular attractions, instead of increasing or decreasing molecular kinetic energies. While a substance is undergoing a change in state, its temperature remains constant.

Given,

Mass is \(94.0\;{\rm{g}}\)

The total heat is calculated as follows,

\(\begin{aligned}{}\Delta {{\rm{H}}_1} &= ({\mathop{\rm mass}\nolimits} )(\Delta {\rm{t}})({\rm{C}})\\\Delta {{\rm{H}}_1} &= 94\;{\rm{g}}\left( {80 - {0^\circ }{\rm{C}}} \right)\left( {4.184\;{\rm{J}}{/^\circ }{\rm{C}} \cdot {\rm{g}}} \right)\\\Delta {{\rm{H}}_1} &= 31463.68\;{\rm{J}}({\rm{ or }})31.46{\rm{KJ}}\end{aligned}\)

\(\begin{aligned}{{}{}}{{\rm{\Delta }}{{\rm{H}}_2} = \frac{{94.0{\rm{g}}}}{{18.02{\rm{g}}/{\rm{mol}}}} \times 6.01{\rm{KJ}}/{\rm{mol}}}\\{{\rm{\Delta }}{{\rm{H}}_2} = 31.351{\rm{KJ}}}\\{{\rm{\Delta }}{{\rm{H}}_3} = (mass)({\rm{\Delta t}})({\rm{C}})}\\{{\rm{\Delta }}{{\rm{H}}_3} = 94{\rm{g}}\left( {30 - {0^ \circ }{\rm{C}}} \right)\left( {2.09{\rm{J}}{/^ \circ }{\rm{C}} \cdot {\rm{g}}} \right)}\\{{\rm{\Delta }}{{\rm{H}}_3} = 5893.8{\rm{J}}({\rm{\;or\;}})5.9{\rm{KJ}}}\end{aligned}\)

The heat is calculated as follows,

Total energy required \( = \Delta {{\rm{H}}_1} + \Delta {{\rm{H}}_2} + \Delta {{\rm{H}}_3}\)

Total energy required \( = 31.46{\rm{KJ}} + 31.351 + 5.9{\rm{KJ}}\)

Total energy required \( = 68.7{\rm{KJ}}\)