A 0.3423 g sample of pentane, C5H12, was burned in a bomb (constant-volume) calorimeter. The temperature of the calorimeter and
the 1.000 kg of water contained therein rose from 20.22°C to 22.82°C. The heat capacity of the calorimeter is 2210J/°C. The specific heat of water is 4.184 J/g·°C. Calculate the heat of combustion of pentane in kJ/g.
1 answer:
Answer:
16.82 kJ/g
Explanation:
The Heat energy generated by the pentance combusion caused the temperature increasing of the caloriemeter and the water inside it.
Heat energy of a substace can be calculated using
Q = mcθ
C= mc
where Q = Heat energy
m =mass
c = specific heat capacity
C = heat capacity
θ =temperature difference
(Temperature difference = final temperature - initial temperature.)
heat absorbed by water = mcθ
= 1 * 4.184 * (22.82-20.22)
= 10.88 J --------------(1)
heat absorbed by calorimeter = Cθ
= 2210 *(22.82-20.22)
= 5746 J ---------------(2)
Combustion energy = heat absorbed by water + heat absorbed by calorimeter
Combustion energy = 10.88+ 5746 = 5756.88 J
= 5.757 kJ
heat of combustion of pentane = Combustion energy / mass
= 5.757 / 0.3423
= 16.82 kJ/g
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- s (displacement) negative, since it is downward: so s = -75.0 m
- u (initial velocity) positive, since it is upward: +15.5 m/s
- a (acceleration) negative, since it is downward: so a= g = -9.8 m/s^2 (acceleration of gravity)
Substituting into the equation,
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b) 10.4 m/s
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where again, we must be careful to the signs of the various quantities:
- u (initial velocity) positive, since it is upward: +15.5 m/s
- a (acceleration) negative, since it is downward: so a = g = -9.8 m/s^2
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c) 4.11 s
In this case, we can use again the equation:
where
s is the displacement of the package
u is the initial velocity
t is the time
a is the acceleration
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