QJEEM24Ph3393: JEE Main 2024 Physics - Class 11 JEEM24Ph3393 Solution: 0.08   kg air is heated at constant volume through 5 ∘ C . …

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:

$$\mathrm{\Delta }U=m{c}_v\mathrm{\Delta }T$$

Where:

• $$\mathrm{\Delta }U$$ 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,
• $$\mathrm{\Delta }T$$ is the change in temperature.

Given that:

• The mass of air, $$m=0.08\text{kg},$$
• The specific heat of air at constant volume, $$c_v=0.17{\text{kcal/kg}}^{\circ }\text{C},$$
• The change in temperature, $$\mathrm{\Delta }T=5^{\circ }\text{C},$$
• And the conversion factor from calories to Joules, $$1\text{cal}=4.18\text{J}.$$

First, convert the specific heat from kcal to Joules:

$$c_v=0.17{\text{kcal/kg}}^{\circ }\text{C}\times 1000\text{cal/kcal}\times 4.18\text{J/cal}=710.6{\text{J/kg}}^{\circ }\text{C}$$

Now, substitute the values into the formula:

$$\mathrm{\Delta }U=0.08\times 710.6\times 5$$

$$\mathrm{\Delta }U=284\text{J}$$

Thus, the change in internal energy is approximately 284 Joules.