Question a540a

Sep 11, 2016

Find all heats according to temperature decrease and phase changes and add them. Answer is:

${Q}_{t o t} = 62.6509 k J$

Explanation:

First you need to understand what phase transitions it faces. From the melting and boiling points you know that:

Transition 1 Temperature cooling : Gas ${92}^{o} C \to {78.4}^{o} C$

Transition 2 Phase change : Condensation at ${78.4}^{o} C$

Transition 3 Temperature cooling : Liquid ${78.4}^{o} C \to - {16}^{o} C$

Since the melting point of $- {114.5}^{o} C$ is not met, the ethanol does not become solid.

The 3 energies you need to estimate are the three transitions mentioned above.

Transition 1

The energy is given as a temperature decrease ΔT_1:

Q_1=n*c_(p(gas))*ΔT=m/(Mr)*c_(p(gas))*ΔT_1

Note: this equation uses mols instead of grams because the heat capacity is given in J/(K*mol).

$M {r}_{{C}_{2} {H}_{5} O H} = 2 \cdot 12 + 1 \cdot 6 + 16 = 46 \frac{g}{m o l}$

${Q}_{1} = \frac{20}{16} \cdot 65.7 \cdot \left(92 - 78.4\right) \frac{g}{\frac{g}{m o l}} \cdot \frac{J}{K \cdot m o l} \cdot K$

${Q}_{1} = = 1116 , 9 J = 1.1169 k J$

Transition 2

The energy is given for the phase transition:

Q_2=ΔH_(vap)*n=ΔH_(vap)*m/(Mr)=38.56*20/16 (kJ)/(mol)*g/(g/(mol))

${Q}_{2} = 48.2 k J$

Transition 3

The energy is given as a temperature decrease ΔT_3:

Q_3=n*c_(p(liq))*ΔT=m/(Mr)*c_(p(liq))*ΔT_3

${Q}_{3} = \frac{20}{16} \cdot 113 \cdot \left(78.4 - \left(- 16\right)\right) \frac{g}{\frac{g}{m o l}} \cdot \frac{J}{K \cdot m o l} \cdot K$

${Q}_{3} = 13334 J = 13.334 k J$

Total energy

The total energy is:

${Q}_{t o t} = {Q}_{1} + {Q}_{2} + {Q}_{3} = 1.1169 k J + 48.2 k J + 13.334 k J$

${Q}_{t o t} = 62.6509 k J$

Water as medium

If the case was water, than the properties would change. Most important properties are boiling and melting points. Water boils at ${100}^{o} C$ so no phase change would occur since the top temperature is ${92}^{o} C$. However, since $- {16}^{o} C$ is lower than the melting point ${0}^{o} C$, water would undergo a phase change as a solid, so enthalpy of fusion should be calculated. Also, water would have different heat capacities. Finally, the molecular weight would be different $\left(M {r}_{{H}_{2} O} = 2 \cdot 1 + 16 = 18 \frac{g}{m o l}\right)$ In short, the heat you would need to find would be equal to:

${Q}_{t o t} = {Q}_{1} + {Q}_{2} + {Q}_{3}$

But their values would be equal to:

Q_1=n*c_(p_(liq))*ΔT_1
(ΔT_1=92-0)

Q_2=ΔH_(fusion)*n

Q_3=n*c_(p_(solid))*ΔT_3
(ΔT_3=0-(-16))#