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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: johnnyjohn993123 on December 22, 2014, 09:13:38 PM

Title: Bomb Calorimetry
Post by: johnnyjohn993123 on December 22, 2014, 09:13:38 PM

A 0.3423 g sample of pentane, CH₁₂, was burned in a bomb 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
2.21 kJ/°C. The heat capacity of water = 4.184 J/g·°C. How much heat was given off during combustion of the sample
of pentane?

my solution:
 q_cal + q_sol'n= - q_reaction

ΔT=0.6°C
m_sol'n=1000g
C=2210J
s=4.184
 q_cal + q_sol'n= - q_reaction

  (C)( ΔT)         +  (  ΔT)(m)(s)     =  - q_reaction

(2210J/°C )(0.6°C)+ (0.6)(1000g)(4.184) =  - q_reaction

 q_reaction = -3836.4 J

I don't know if this is correct . Do I have to disregard the mass of pentane?

Title: Re: Bomb Calorimetry
Post by: Borek on December 23, 2014, 05:17:44 AM

I have not checked numbers, but the logic looks OK.

Yes, you can ignore pentane presence. That's an approximation only, but a reasonably good one  (compare mass of pentane with mass of water - that's the order of magnitude of the error introduced by the approximation).

Pentane is C₅H₁₂, but I guess it is just a typo.

Title: Re: Bomb Calorimetry
Post by: mjc123 on December 23, 2014, 06:28:21 PM

ΔT is not 0.6°C, is it?