0.08 kg of air is heated at constant volume through 5°C. The specific heat of air at constant volume is 0.17 kcal/kg°C and $ J = 4.18 $ joule/cal. The change in its internal energy is approximately:

Question

Given $C_p = a + bT$
$a = 19.5$, $b = 0.042$, $n = 1$ mole
Temperature changes from $27^\circ C$ to $327^\circ C$.
Find $\Delta H$.

Answer 1

The change in internal energy for air heated at constant volume is determined using the formula:

$\Delta U = m \cdot C_v \cdot \Delta T$

Initially, the change in internal energy is calculated in kcal:

$\Delta U = 0.08 \, \text{kg} \times 0.17 \, \text{kcal/kg°C} \times 5\,°C$

$\Delta U = 0.068\, \text{kcal}$

Next, convert kcal to joules using the conversion factor $1\, \text{kcal} = 4184\, \text{joules}$:

$\Delta U = 0.068\, \text{kcal} \times 4184\, \text{J/kcal}$

$\Delta U = 284.512\, \text{J}$

The calculated change in internal energy is approximately 284 J.

The correct answer is 284 J, as other options do not align with this result.

0.08 kg of air is heated at constant volume through 5°C. The specific heat of air at constant volume is 0.17 kcal/kg°C and \( J = 4.18 \) joule/cal. The change in its internal energy is approximately: