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Q.

2 kg of air is heated at constant volume.The temperature of air is increased from 293 K to 313 K.If the specific heat of air at constant volume is 0.718 kJ/kg K, the amount of heat absorbed in kJ and kcal is, (J = 4.2/kcal)

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a

28.72 KJ , 6.838 kcal

b

32.36 KJ,7.8 kJ

c

61.5 Kcal ,14.68 kJ

d

111.5 Kcal,28.72 kJ

answer is A.

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Detailed Solution

$\begin{array}{l} m = 2 kg; dt = t_{2} - t_{1} = 313 - 293 = 20 k \\ C_{v} = 0 .718 kJ / kg - k; (dQ)_{v} = ? \\ C_{v} \frac{1}{m} \frac{{(d_{Q})}_{v}}{dt} \Rightarrow (dQ)_{v} = {mc}_{v} dt = 2 \times 0 .718 \times 20 \\ 28 .72 KJ = \frac{28 .72}{4 .2} = 6 .838 kcal \end{array}$

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