Respuesta :
Answer: The amount of heat needed to melt the given amount of octane is 84.6 kJ
Explanation:
We know:
Boiling point of Octane = [tex]125.6^oC[/tex]
Few processes involved are:
(1): [tex]C_8H_{18} (s) (-57^oC, 219K) \rightleftharpoons C_8H_{18}(s) (-57^oC, 219K[/tex]
(2): [tex]C_8H_{18}(l) (-57^oC, 219K) \rightleftharpoons C_8H_{18}(l) (99.2^oC,372.2K)[/tex]
Calculating the heat absorbed for the process having same temperature:
[tex]q=n\times \Delta H_{(f)}[/tex] ......(i)
where,
q is the amount of heat absorbed, n is the moles of sample and [tex]\Delta H_{(f)}[/tex] is the enthalpy of fusion
Calculating the heat released for the process having different temperature:
[tex]q=n\times C_{l}\times (T_2-T_1)[/tex] ......(ii)
where,
n = moles of sample
[tex]C_{l}[/tex] = specific heat of liquid
[tex]T_2\text{ and }T_1[/tex] are final and initial temperatures respectively
The number of moles is defined as the ratio of the mass of a substance to its molar mass.
The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(3)
Given mass of octane = 160. g
Molar mass of octane = 114.23 g/mol
Plugging values in equation 3:
[tex]\text{Moles of octane }=\frac{160.g}{114.23g/mol}=1.40 mol[/tex]
- For process 1:
We are given:
[tex]n=1.40mol\\\Delta H_{fusion}=20.740 kJ/mol[/tex]
Putting values in equation (i), we get:
[tex]q_1=1.40mol\times 20.470kJ/mol\\\\q_1=28.658kJ[/tex]
- For process 2:
We are given:
[tex]n=1.40mol\\C=255.68J/mol^oC\\T_2=99.2^oC\\T_1=-57^oC[/tex]
Putting values in equation (ii), we get:
[tex]q_2=1.40mol\times 255.68J/mol^oC\times (99.2-(-57))\\\\q_2=55912.10J=55.912kJ[/tex]
Calculating the total amount of heat released:
[tex]Q=q_1+q_2[/tex]
[tex]Q=[(28.658)+(55.912)]kJ=84.6kJ[/tex]
Hence, the amount of heat needed to melt the given amount of octane is 84.6 kJ
