Question 201: Thermodynamics & KTG

Question

$0.08 \mathrm{~kg}$ air is heated at constant volume through $5^{\circ} \mathrm{C}$. The specific heat of air at constant volume is $0.17 \mathrm{~kcal} / \mathrm{kg}^{\circ} \mathrm{C}$ and $\mathrm{J}=4.18$ joule/$\mathrm{~cal}$. The change in its internal energy is approximately.

Accepted Answer

To find the change in the internal energy of air when it is heated at constant volume, we use the formula for heat transfer at constant volume, which is given by:
$\Delta U = m c_v \Delta T$
Where:
$\Delta U$ is the change in internal energy,
$m$ is the mass of the substance (in this case, air),
$c_v$ is the specific heat at constant volume,
$\Delta T$ is the change in temperature.
Given that:
The mass of air, $m = 0.08 \, 	ext{kg},$
The specific heat of air at constant volume, $c_v = 0.17 \, 	ext{kcal/kg}^{\circ}	ext{C},$
The change in temperature, $\Delta T = 5^{\circ} 	ext{C},$
And the conversion factor from calories to Joules, $1 \, 	ext{cal} = 4.18 \, 	ext{J}.$
First, convert the specific heat from kcal to Joules:
$c_v = 0.17 \, 	ext{kcal/kg}^{\circ}	ext{C} 	imes 1000 \, 	ext{cal/kcal} 	imes 4.18 \, 	ext{J/cal} = 710.6 \, 	ext{J/kg}^{\circ}	ext{C}$
Now, substitute the values into the formula:
$\Delta U = 0.08 	imes 710.6 	imes 5$
$\Delta U = 284 \, 	ext{J}$
Thus, the change in internal energy is approximately 284 Joules.

Answer

318 J

Answer

298 J

Answer

284 J

Answer

142 J

Thermodynamics & KTG

0.08 \mathrm{~kg}$ air is heated at constant volume through $5^{\circ} \mathrm{C}$. The specific heat of air at constant volume is $0.17 \mathrm{~kcal} / \mathrm{kg}^{\circ} \mathrm{C}$ and $\mathrm{J}=4.18$ joule/$\mathrm{~cal}$. The change in its internal energy is approximately.

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