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yan [13]
3 years ago
12

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

.2 K. What is the enthalpy of combustion of octane? The specific heat capacity of water is 4.18 J/K g.
A. -1226 kJ/mol
B. -5448 kJ/mol
C. 293.25 kJ/mol
D. 1226 kJ/mol
Physics
1 answer:
arsen [322]3 years ago
5 0

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

q=m\times c\times (T_2-T_1)

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = 4.18J/g.K

T_1 = initial temperature of water = 293.0 K

T_2 = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

q=250g\times 4.18J/g.K\times (371.2-293.0)K

q=81719J

Now we have to calculate the enthalpy of combustion of octane.

\Delta H=\frac{q}{n}

where,

\Delta H = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

\Delta H=\frac{-81719J}{0.015mole}

\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

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