Question

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:

• A. 318 J
• B. 298 J
• C. 284 J
• D. 142 J

Answer 1

The Correct Option is C

To find the change in internal energy for air when heated at constant volume, we can use the formula for the change in internal energy at constant volume: 

\(\Delta U = m \cdot C_v \cdot \Delta T\)

• \(m\) is the mass of the air: \(0.08 \, \text{kg}\)
• \(C_v\) is the specific heat at constant volume (in kcal/kg°C): \(0.17 \, \text{kcal/kg°C}\)
• \(\Delta T\) is the change in temperature: \(5\,°C\)

First, let's calculate the change in internal energy in kcal:

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

\(\Delta U = 0.068\, \text{kcal}\)

Now, convert kcal to joules using the conversion factor \(1\, \text{kcal} = 4184\, \text{joules}\):

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

\(\Delta U = 284.512\, \text{J}\)

Therefore, the change in internal energy is approximately 284 J.

The correct answer is therefore 284 J. The other options do not match the calculated value.

Answer 2

Since the process is at constant volume, the change in internal energy \(\Delta U\) is given by:

\[\Delta U = ms\Delta T\]

where \(m = 0.08 \, \text{kg}\), \(s = 0.17 \, \text{kcal/kg}^\circ\text{C}\), and \(\Delta T = 5^\circ \text{C}\).

Convert \(s\) from kcal to joules:

\[s = 0.17 \times 1000 \times 4.18 \, \text{J/kg}^\circ\text{C}\]

Then,

\[\Delta U = 0.08 \times (0.17 \times 1000 \times 4.18) \times 5 \approx 284 \, \text{J}\]