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Calorimeter

The thermal c...

Question

The thermal capacity of calorimeter system is $17.7 k J$ $K^{- 1}$ . ( $R = 8.313 {m o l}^{- 1} K^{- 1}$ ) (only magnitude in nearest integer in kJ/mol)

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Solution

At constant volume
$Δ H = Δ U$
Rise in temperature of the system = ${0.5}^{o} C = 0.5 K$
Total heat released $= - C_{p} Δ T = - 17.7 \frac{k J}{K} \times 0.5 K = - 8.85 k J$
0.16 g of methane $= \frac{0.16}{16}$ mole $= 0.01$ mole of methane (molecular weight of methane =16 g)
Heat of combustion $= \frac{t o t a h e a t r e l e a s e d}{n u m b e r o f m o l e s} = \frac{- 8.85}{0.01} = - 885 k J m o l^{- 1}$
Answer = 885

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