Respuesta :
Answer:
The heat required to convert 75.0 g of [tex]C_2Cl_3F_3[/tex] from a liquid at 11.50 °C to a gas at 80.30 °C is 15.103 kiloJoules.
Explanation:
1) Heat required to raise the temperature of fluorocarbon compound from 11.50°C to 47.6 °C= [tex]Q_1[/tex]
[tex]C_2Cl_3F_3(l,284.65 K)\rightarrow C_2Cl_3F_3(l,320.75 K)[/tex]
Mass of fluorocarbon compound = m = 75.0 g
Specific heat of fluorocarbon compound ,c= 0.91 J/gK
Change in temperature = [tex]\Delta T=320.75 K-284.65 K=36.1 K[/tex]
[tex]Q_1=m\times c\times \Delta T[/tex]
[tex]=75.0 g\times 0.91 J/g K\times 36.1 K=2,463.82 J[/tex]
2) Heat required to raise the change the state of of fluorocarbon compound from liquid to gas= [tex]Q_2[/tex]
[tex]C_2Cl_3F_3(l)\rightarrow C_2Cl_3F_3(g)[/tex]
Mass of fluorocarbon compound = m = 75.0 g
Moles of fluorocarbon compound = [tex]\frac{75.0 g}{187.5 g/mol}=0.4 mol[/tex]
The heat of vaporization for the compound = [tex]\Delta H_{vap}=27.49 kJ/mol[/tex]
[tex]Q_2= \Delta H_{vap}\times 0.4 mol=27.49 kJ/mol\times 0.4 mol=10.996 kJ[/tex]
[tex]Q_2=10.996 kJ = 10,996 J[/tex] (1 kJ= 1000 J)
3) Heat required to raise the temperature of fluorocarbon compound from 47.6 °C to 80.30°C= [tex]Q_3[/tex]
[tex]C_2Cl_3F_3(g,320.75 K)\rightarrow C_2Cl_3F_3(g,353.45 K)[/tex]
Mass of fluorocarbon compound = m = 75.0 g
Specific heat of fluorocarbon compound ,c'= 0.67 J/gK
Change in temperature = [tex]\Delta T'=353.45 K-320.75 K=32.7 K[/tex]
[tex]Q_3=m\times c'\times \Delta T'[/tex]
[tex]=75.0 g\times 0.67 J/g K\times 32.7 K=1,643.18 J[/tex]
The heat required to convert 75.0 g of [tex]C_2Cl_3F_3[/tex] from a liquid at 11.50 °C to a gas at 80.30 °C:
[tex]Q_1+Q_2+Q_3=2,463.82 J+10,996 J+1,643.18 J=15,103 J= 15.103 kJ[/tex]
The heat required to convert 75.0 g of C2Cl3F3 from a liquid at 11.50 °C to gas at 80.30 °C is mathematically given as
Q = 7916.59 J
What is the heat required to convert 75.0 g of C2Cl3F3 from a liquid at 11.50 °C to gas at 80.30 °C?
Question Parameter(s):
has a normal boiling point of 47.6 °C.
The specific heats of C2Cl3F3(l) and C2Cl3F3(g) are 0.91 J/g.K and 0.67 J/g.K, respectively
The heat of vaporization for the compound is 27.49 kJ/mol.
Generally, the equation for the Total Heat is mathematically given as
Q = Q1+ Q2+ Q3
Therefore
Q1 = m * cL * dTL
Q2 = m * Lv
Q3= m * cG * dTG
Where latent heat of vaporization is given as
Lv / M = 27.49 kJ/mol / 187.5 gr/mol
Lv / M =0.1466 kJ/g
Hence
Q1 = 38 g * 0.91 * ( 47.6°C - 12.3°C)
Q1= 1186.09 J
Q2= 38 g * 0.1466 kJ/g * 1000J/kJ
Q2= 5570.8 J
Q3= 38 g * 0.67 J/gK * (93.15°C - 47.6°C )
Q3= 1159.70 J
In conclusion,
Q =1186.09 J + 5570.8 J + 1159.70 J
Q = 7916.59 J
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