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Question: \(2kg\) of air is heated at constant volume. The temperature of the air has increased from \(293K\) ...

2kg2kg of air is heated at constant volume. The temperature of the air has increased from 293K293K to 313K313K. If the specific heat of air at constant volume is 0.718KJ/kgK0.718KJ/kgK, the amount of heat absorbed in kJkJ and kcalkcal is (J=4.2/kcal)\left( {J = 4.2/kcal} \right)
A. 6.838kJ6.838kJ
B. 7.8kJ7.8kJ
C. 14.68kJ14.68kJ
D. 28.72kJ28.72kJ

Explanation

Solution

To solve this question, we need to use the formula for the heat absorbed in the constant volume process. Substituting the values of the mass of air, specific heat at constant volume, and the change in temperature from the information given in the question, we can get the final answer.

Formula used: The formula used to solve this question is given by
Q=mCVΔTQ = m{C_V}\Delta T, here QQ is the amount of heat absorbed or rejected by mm mass of a substance having the specific heat at constant volume of CV{C_V} when it is subjected to a change in the temperature of ΔT\Delta T.

Complete step by step answer:
According to the question, we have a constant volume process in which a given amount of air is heated. We know that the heat exchanged by a substance in a constant volume process is given by
Q=mCVΔTQ = m{C_V}\Delta T
Q=mCV(T2T1)\Rightarrow Q = m{C_V}\left( {{T_2} - {T_1}} \right)
Now, according to the question we have the mass of air equal to 2kg2kg, the specific heat capacity at constant volume of air is equal to 0.718KJ/kgK0.718KJ/kgK. Also, the initial temperature of air is equal to 293K293K and its final temperature is equal to 313K313K. Therefore substituting m=2kgm = 2kg, CV=0.718KJ/kgK{C_V} = 0.718KJ/kgK, T1=293K{T_1} = 293K, and T2=313K{T_2} = 313K in the above equation, we get
Q=2×0.718×(313293)\Rightarrow Q = 2 \times 0.718 \times \left( {313 - 293} \right)
On solving we get
Q=28.72kJ\Rightarrow Q = 28.72kJ
Thus, the heat absorbed by the air is equal to 28.72kJ28.72kJ.
Hence, the correct answer is option D.

Note: Do not convert the given value of the specific heat capacity at constant volume into its SI unit, that is, from KJ/kgKKJ/kgK to J/kgKJ/kgK. This is because we are asked to find out the amount of heat absorbed in kilo-joules. So directly substituting the value of the heat capacity in the formula, we got the heat absorbed in kilo-joules.