First slide

Thermodynamic properties

Question

Heat capacity of water is 18 cal-degree^-1-mol^-1. The quantity of heat needed to rise temperature of 18g water by 0.2^oC is X cal. Then amount of CH₄(g) to be burnt to produce X cal heat is
(CH₄+2O₂ →CO₂+2H₂O, ΔH= –200K.Cal)

Moderate

Solution

q =m C_pΔT
$=\frac{{18}}{{18}}\; \times \;18\; \times \;0.2\$

= 3.6 cal
16gm -------200Kcal
? -------3.6 x 10^-3 Kcal

$\large =\16 \times \frac{{3.6}}{{2000}} \times {10^{ - 3}}\$

= 288 x 10^-6gm
= 0.288 mg

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