# From the data below, calculate the total heat (in J) needed to convert 0.670 mol of gaseous...

## Question:

From the data below, calculate the total heat (in J) needed to convert 0.670 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 H vap: 40.5 kJ/mol

c ethanol( g ): 1.43 J/g *C

c ethanol( l ): 2.45 J/g *C

## Heat Transfer:

From a system, the heat transfer is occurring in the following modes.

1. Heat transfer due to radiation
2. Heat transfer due to convection
3. Heat transfer due to conduction

Heat transfer can be calculated by using the following expression.

{eq}Q = mc\left( {{T_f} - {T_i}} \right) {/eq}

Here, the heat transfer due to variation in the temperature is {eq}Q {/eq}, the mass is m, the specific heat capacity is c, the initial temperature is {eq}{T_i} {/eq} and the final temperature of the system is {eq}{T_f} {/eq}

## Answer and Explanation:

Given data

• The number of moles of gaseous ethanol is {eq}n = 0.67\;{\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( {0.67\;{\rm{mol}}} \right) \times \left( {4{\rm{6}}\;{\rm{g/mol}}} \right)\\ &= 30.82\;{\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( {30.82\;{\rm{g}}} \right)\left( {1.43\;{\rm{J/g^\circ C}}} \right)\left( {300{\rm{^\circ C}} - 78.5{\rm{^\circ C}}} \right) + \left( {0.67\;{\rm{mol}}} \right)\left( {40.5\;{\rm{kJ/mol}}} \right) + \left( {30.82\;{\rm{g}}} \right)\left( {2.45\;{\rm{J/g^\circ C}}} \right)\left( {78.5{\rm{^\circ C}} - 25{\rm{^\circ C}}} \right)\\ &= 9762.08\;{\rm{J}} + \left( {27.135\;{\rm{kJ}}} \right)\left( {\dfrac{{1000\;{\rm{J}}}}{{1\;{\rm{kJ}}}}} \right) + 4039.73\;{\rm{J}}\\ &= 9762.08\;{\rm{J}} + 27135\;{\rm{J}} + 4039.73\;{\rm{J}}\\ &= 40936.81\;{\rm{J}} \end{align*} {/eq}

Thus, the total heat needed is 40936.81 J.

#### Learn more about this topic:

Thermal Expansion & Heat Transfer

from High School Physics: Help and Review

Chapter 17 / Lesson 12
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