# From the data below, calculate the total heat (in J) needed to convert 0.261 mol of gaseous...

## Question:

From the data below, calculate the total heat (in J) needed to convert 0.261 mol of gaseous ethanol at 300.0 ° C and 1 atm to liquid ethanol at 25.0 ° C and 1 atm:

b.p. at 1 atm: 78.5 ° C

{eq}Δ H ° _{vap} {/eq} : 40.5 kJ/mol

{eq}c _{ethanol(g)} {/eq} : 1.43 J/g · ° C

{eq}c _{ethanol(l)} {/eq} : 2.45 J/g · ° C

## Heat Transfer:

The term used in thermal engineering to describe the process of the use of thermal energy is represented by heat transfer. Heat transfer is also utilized to describe the procedure of the conversion of thermal energy.

Given data

• The number of moles of gaseous ethanol is {eq}n = 0.261\;{\rm{mol}} {/eq}
• The temperature of the gaseous ethanol is {eq}{T_1} = 300{\rm{^\circ C}} {/eq}
• The pressure of the gaseous ethanol is {eq}{P_1} = 1\;{\rm{atm}} {/eq}
• The temperature of the liquid ethanol is {eq}{T_2} = 25{\rm{^\circ C}} {/eq}
• The pressure of the liquid ethanol is {eq}{P_2} = 1\;{\rm{atm}} {/eq}
• The temperature is of b.p is {eq}{T_3} = 78.5{\rm{^\circ C}} {/eq}
• The change in enthalpy is {eq}\Delta H = 40.5\;{\rm{kJ/mol}} {/eq}
• The heat capacity of gaseous ethanol is {eq}{C_g} = 1.43\;{\rm{J/g^\circ C}} {/eq}
• The heat capacity of liquid ethanol is {eq}{C_l} = 2.45\;{\rm{J/g^\circ C}} {/eq}

Note- The molar mass of the ethanol is M = 46 g/mol

The mass of the ethanol is calculated as,

{eq}\begin{align*} m &= n \times M\\ &= \left( {0.261\;{\rm{mol}}} \right) \times \left( {4{\rm{6}}\;{\rm{g/mol}}} \right)\\ &= 12.006\;{\rm{g}} \end{align*} {/eq}

The expression of the total heat needed is calculated as

{eq}\begin{align*} Q &= {Q_1} + {Q_2} + {Q_3}\\ &= m{C_g}\left( {{T_1} - {T_3}} \right) + n\Delta H + m{C_l}\left( {{T_3} - {T_2}} \right)\\ &= \left( {12.006\;{\rm{g}}} \right)\left( {1.43\;{\rm{J/g^\circ C}}} \right)\left( {300{\rm{^\circ C}} - 78.5{\rm{^\circ C}}} \right) + \left( {0.261\;{\rm{mol}}} \right)\left( {40.5\;{\rm{kJ/mol}}} \right) + \left( {12.006\;{\rm{g}}} \right)\left( {2.45\;{\rm{J/g^\circ C}}} \right)\left( {78.5{\rm{^\circ C}} - 25{\rm{^\circ C}}} \right)\\ &= 3802.84\;{\rm{J}} + \left( {10.5705\;{\rm{kJ}}} \right)\left( {\dfrac{{1000\;{\rm{J}}}}{{1\;{\rm{kJ}}}}} \right) + 1573.686\;{\rm{J}}\\ &= 3802.84\;{\rm{J}} + 10570.5\;{\rm{J}} + 1573.686\;{\rm{J}}\\ &= 15947.026\;{\rm{J}} \end{align*} {/eq}

Thus, the total heat needed is 15947.026 J.