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At 1 atm, how much energy is required to heat 55.0g of H2O(s) at -22.0 ^degree...

At 1 atm, how much energy is required to heat 55.0g of H2O(s) at -22.0 ^degree C to H2O(g) at 123.0^degree C? Please answer in kJ & explain in detail
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Answer #1
Concepts and reason

When heat is gained by the substance, a certain amount of temperature will change. For every material, the relationship between temperature and heat energy is different. The value of specific heat describes the relationship between temperature and heat energy of each and every material.

The amount of energy required to melt the solid is calculated by the enthalpy of fusion.

The amount of energy required to change 1 gram of liquid to gas is represented as heat of vaporization.

Fundamentals

The energy required for melting can be calculated as follows:

q=mHfHf=enthalpyoffusionm=mass\begin{array}{l}\\{\rm{q}}\,{\rm{ = }}\,\,{\rm{m}}\,{{\rm{H}}_{\rm{f}}}\\\\{{\rm{H}}_{\rm{f}}}{\rm{ = }}\,{\rm{enthalpy of fusion}}\\\\{\rm{ m}}\,{\rm{ = }}\,{\rm{mass}}\\\end{array}

The energy required for vaporizing can be calculated as follows:

q=mHvHv=enthalpyofvaporizationm=mass\begin{array}{l}\\{\rm{q}}\,{\rm{ = }}\,\,{\rm{m}}\,{{\rm{H}}_v}\\\\{{\rm{H}}_v}{\rm{ = }}\,{\rm{enthalpy of vaporization}}\\\\{\rm{ m}}\,{\rm{ = }}\,{\rm{mass}}\\\end{array}

The energy with respect to specific heat can be calculated as follows:

q=mcΔTm=massc=specificheatΔT=changeintemperature\begin{array}{l}\\\,{\rm{q}}\,\,{\rm{ = }}\,\,{\rm{mc\Delta T}}\\\\{\rm{m}}\,\,{\rm{ = }}\,\,{\rm{mass}}\\\\{\rm{c}}\,\,\,\,{\rm{ = }}\,\,{\rm{specific}}\,{\rm{heat}}\\\\{\rm{\Delta T = }}\,{\rm{change}}\,\,{\rm{in}}\,\,{\rm{temperature}}\\\end{array}

Heattomovethesolidwaterfrom22.0to0.00C:Specificheatofsolidwater=2.135J/g0Cmassofsolidwater=55.0ginitialtemperature=220Cfinaltemperature=0.00C\begin{array}{l}\\{\rm{Heat to move the solid water from - 22}}{\rm{.0 to 0}}{\rm{.}}{{\rm{0}}^{\rm{0}}}{\rm{C:}}\\\\{\rm{Specific}}\,\,{\rm{heat}}\,\,{\rm{of}}\,\,{\rm{solid water}}\,\,{\rm{ = }}\,\,{\rm{2}}{\rm{.135}}\,{\rm{J/}}{{\rm{g}}^{\rm{0}}}{\rm{C}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{mass}}\,\,{\rm{of}}\,\,{\rm{solid water}}\,\,{\rm{ = }}\,\,{\rm{55}}{\rm{.0g}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{initial}}\,\,{\rm{temperature}}\,\,{\rm{ = }}\,\,{\rm{ - 2}}{{\rm{2}}^{\rm{0}}}{\rm{C}}\\\\ & \,\,\,\,\,\,{\rm{final}}\,\,{\rm{temperature}}\,\,\,{\rm{ = }}\,\,{\rm{0}}{\rm{.}}{{\rm{0}}^{\rm{0}}}{\rm{C}}\\\end{array}

q=mCΔT=55.0g×2.135Jg0C×(0.0(22.0))0Cq=2583J\begin{array}{l}\\{\rm{q}}\,\,{\rm{ = }}\,\,{\rm{mC\Delta T}}\\\\\,\,\,\,\,\,\,{\rm{ = }}\,{\rm{55}}{\rm{.0g}}\,\,{\rm{ \times }}\,\,{\rm{2}}{\rm{.135}}\,\frac{{\rm{J}}}{{{\rm{g}}{\,^{\rm{0}}}{\rm{C}}\,}}{\rm{ \times }}\,{\left( {{\rm{0}}{\rm{.0}}\,{\rm{ - }}\,\left( {{\rm{ - 22}}{\rm{.0}}} \right)} \right)^{\rm{0}}}{\rm{C}}\\\\{\rm{q}}\,\,{\rm{ = }}\,{\rm{2583}}\,\,{\rm{J}}\\\end{array}

Heatoffusionofwater=334Jgq=mHfq=55.0g×334Jgq=18370J\begin{array}{l}\\{\rm{Heat}}\,\,{\rm{of}}\,\,{\rm{fusion}}\,\,{\rm{of}}\,\,{\rm{water}}\,\,{\rm{ = }}\,\,{\rm{334}}\,\frac{{\rm{J}}}{{\rm{g}}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,{\rm{q}}\,\,{\rm{ = }}\,{\rm{m}}{{\rm{H}}_{\rm{f}}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,{\rm{q}}\,\,{\rm{ = }}\,{\rm{55}}{\rm{.0g}}\,\,{\rm{ \times 334}}\,\frac{{\rm{J}}}{{\rm{g}}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,{\rm{q}}\,\,{\rm{ = }}\,\,{\rm{18370}}\,{\rm{J}}\\\end{array}

The boiling point of water is 1000C.

Heattomovetheliquidwaterfrom0.0to1000CSpecificheatofliquidwater=4.184Jg0Cmassofliquidwater=55.0ginitialtemperature=0.00Cfinaltemperature=1000C\begin{array}{l}\\{\rm{Heat}}\,{\rm{to}}\,\,{\rm{move}}\,\,{\rm{the}}\,\,{\rm{liquid}}\,\,{\rm{water}}\,\,{\rm{from}}\,\,{\rm{0}}{\rm{.0}}\,{\rm{to}}\,{\rm{10}}{{\rm{0}}^{\rm{0}}}{\rm{C}}\\\\{\rm{Specific}}\,\,{\rm{heat}}\,\,{\rm{of}}\,{\rm{liquid}}\,\,{\rm{water}}\,\,{\rm{ = }}\,\,{\rm{4}}{\rm{.184}}\,\frac{{\rm{J}}}{{{\rm{g}}{\,^{\rm{0}}}{\rm{C}}}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{mass}}\,\,{\rm{of}}\,{\rm{liquid}}\,\,{\rm{water}}\,\,{\rm{ = }}\,{\rm{55}}{\rm{.0g}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{initial}}\,\,{\rm{temperature}}\,\,{\rm{ = }}\,\,{\rm{0}}{\rm{.0}}{\,^{\rm{0}}}{\rm{C}}\\\\ & \,\,\,\,\,\,\,{\rm{final}}\,\,\,{\rm{temperature}}\,\,{\rm{ = }}\,\,{\rm{10}}{{\rm{0}}^{\rm{0}}}{\rm{C}}\\\end{array}

q=mCΔT=55.0g×4.184Jg0C×(1000)0Cq=23012J\begin{array}{l}\\\,\,\,\,\,\,\,\,\,\,\,{\rm{q}}\,\,{\rm{ = }}\,\,{\rm{mC\Delta T}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{ = }}\,\,{\rm{55}}{\rm{.0g}}\,{\rm{ \times }}\,{\rm{4}}{\rm{.184}}\,\frac{{\rm{J}}}{{{\rm{g}}{\,^{\rm{0}}}{\rm{C}}}}{\rm{ \times }}\,{\left( {{\rm{100}}\,{\rm{ - 0}}} \right)^{\rm{0}}}{\rm{C}}\\\\\,\,\,\,\,\,\,\,\,\,\,{\rm{q}}\,\,\,{\rm{ = }}\,\,{\rm{23012}}\,{\rm{J}}\\\end{array}

Heatofvaporizationofwater=2257Jgmassofwater=55.0g\begin{array}{l}\\{\rm{Heat}}\,\,{\rm{of}}\,\,{\rm{vaporization}}\,\,{\rm{of}}\,\,{\rm{water}}\,\,{\rm{ = }}\,{\rm{2257}}\,\frac{{\rm{J}}}{{\rm{g}}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{mass}}\,{\rm{of}}\,{\rm{water}}\,\,{\rm{ = }}\,\,{\rm{55}}{\rm{.0g}}\\\end{array}

q=mHV=55.0g×2257Jgq=124135J\begin{array}{l}\\{\rm{q}}\,\,{\rm{ = }}\,\,{\rm{m}}{{\rm{H}}_{\rm{V}}}\\\\\,\,\,\,\,\,\,\,{\rm{ = }}\,\,{\rm{55}}{\rm{.0g}}\,{\rm{ \times }}\,{\rm{2257}}\,\frac{{\rm{J}}}{{\rm{g}}}\\\\\,\,\,\,\,{\rm{q}}\,{\rm{ = }}\,\,{\rm{124135}}\,\,{\rm{J}}\\\end{array}

Specificheatofgaseouswater=2.009Jg0Cmassofgaseouswater=55.0g\begin{array}{l}\\\,{\rm{Specific}}\,\,{\rm{heat}}\,\,{\rm{of}}\,\,{\rm{gaseous}}\,\,{\rm{water}}\,\,{\rm{ = }}\,\,{\rm{2}}{\rm{.009}}\,\frac{{\rm{J}}}{{{{\rm{g}}^{\rm{0}}}{\rm{C}}}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{mass}}\,\,{\rm{of}}\,\,{\rm{gaseous}}\,\,{\rm{water}}\,\,{\rm{ = }}\,\,{\rm{55}}{\rm{.0g}}\\\end{array}

Theheattomovethegaseouswaterfrom100.00Cto123.00CInitialtemperature=1000CFinaltemperature=1230Cq=mCΔT\begin{array}{l}\\\,{\rm{The}}\,\,{\rm{heat}}\,\,{\rm{to}}\,\,{\rm{move}}\,\,{\rm{the}}\,\,{\rm{gaseous}}\,\,{\rm{water}}\,\,{\rm{from}}\,{\rm{100}}{\rm{.}}{{\rm{0}}^{\rm{0}}}{\rm{C}}\,\,{\rm{to}}\,{\rm{123}}{\rm{.}}{{\rm{0}}^{\rm{0}}}{\rm{C}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{Initial}}\,\,{\rm{temperature}}\,\,{\rm{ = }}\,\,{\rm{100}}{\,^{\rm{0}}}{\rm{C}}\,\,\,\,\,\,\,\\\\ & \,\,\,\,\,\,\,\,\,\,\,\,{\rm{Final}}\,\,\,{\rm{temperature}}\,\,\,{\rm{ = }}\,{\rm{123}}{\,^{\rm{0}}}{\rm{C}}\,\,\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{q}}\,\,{\rm{ = }}\,\,{\rm{mC\Delta T}}\\\end{array}

=55.0g×2.009Jg0C×(123100)0Cq=2541J\begin{array}{l}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{ = }}\,\,{\rm{55}}{\rm{.0}}\,{\rm{g}}\,{\rm{ \times }}\,\,{\rm{2}}{\rm{.009}}\,\frac{{\rm{J}}}{{{{\rm{g}}^{\rm{0}}}{\rm{C}}\,}}\,{\rm{ \times }}\,{\left( {{\rm{123}}\,{\rm{ - 100}}} \right)^{\rm{0}}}{\rm{C}}\\\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{q}}\,\,\,{\rm{ = }}\,\,{\rm{2541}}\,{\rm{J}}\\\end{array}

Totalenergy:2583J+18370J+23012J+124135J+2541J=170641J170641J×1kJ1000J=171kJ\begin{array}{l}\\{\rm{Total}}\,\,{\rm{energy:}}\\\\{\rm{2583}}\,{\rm{J}}\,\,{\rm{ + }}\,\,{\rm{18370}}\,{\rm{J}}\,\,{\rm{ + }}\,{\rm{23012}}\,{\rm{J}}\,{\rm{ + }}\,{\rm{124135}}\,{\rm{J}}\,{\rm{ + }}\,{\rm{2541}}\,{\rm{J}}\,\,{\rm{ = }}\,{\rm{170641}}\,{\rm{J}}\,\,\\\\{\rm{170641}}\,\,{\rm{J}}\,\,{\rm{ \times }}\,\,\frac{{{\rm{1}}\,{\rm{kJ}}}}{{{\rm{1000J}}}}\,\,{\rm{ = }}\,\,{\rm{171}}\,{\rm{kJ}}\\\end{array}

Ans:

Heat energy required to heat the solid water to gaseous water is 171kJ.

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