$0.08 \mathrm{~kg}$ 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.

<p>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:</p> <p>$\Delta U = m c_v \Delta T$</p> <p>Where:</p> <ul> <li>$\Delta U$ is the change in internal energy,</li> <li>$m$ is the mass of the substance (in this case, air),</li> <li>$c_v$ is the specific heat at constant volume,</li> <li>$\Delta T$ is the change in temperature.</li> </ul> <p>Given that:</p> <ul> <li>The mass of air, $m = 0.08 \, \text{kg},$</li> <li>The specific heat of air at constant volume, $c_v = 0.17 \, \text{kcal/kg}^{\circ}\text{C},$</li> <li>The change in temperature, $\Delta T = 5^{\circ} \text{C},$</li> <li>And the conversion factor from calories to Joules, $1 \, \text{cal} = 4.18 \, \text{J}.$</li> </ul> <p>First, convert the specific heat from kcal to Joules:</p> <p>$$c_v = 0.17 \, \text{kcal/kg}^{\circ}\text{C} \times 1000 \, \text{cal/kcal} \times 4.18 \, \text{J/cal} = 710.6 \, \text{J/kg}^{\circ}\text{C}$$</p> <p>Now, substitute the values into the formula:</p> <p>$\Delta U = 0.08 \times 710.6 \times 5$</p> <p>$\Delta U = 284 \, \text{J}$</p> <p>Thus, the change in internal energy is approximately 284 Joules.</p>