A calorimeter of water equivalent 20 g contains 180 g of water at 25°C. 'm' grams of steam at 100°C is mixed in it till the temperature of the mixture is 31°C. The value of 'm' is close to (Latent heat of water = 540 cal g$$^{-1}$$, specific heat of water = 1 cal g$$^{-1}$$°C$$^{-1}$$)

We have a calorimeter whose water equivalent is 20 g. This means it behaves exactly as if it contained 20 g of water: its thermal capacity is $$20 \text{ cal }{}^\circ\text{C}^{-1}$$. Inside it are $$180 \text{ g}$$ of actual water. The initial temperature of everything inside the calorimeter is $$25^\circ\text{C}$$.

Now, let $$m$$ grams of dry steam at $$100^\circ\text{C}$$ be passed into the calorimeter. After all the steam has condensed and equilibrium is reached, the common temperature of the mixture is $$31^\circ\text{C}$$. We must find the numerical value of $$m$$.

Principle used: Heat lost = Heat gained. No heat is lost to the surroundings.

Heat gained is by the existing water plus the calorimeter itself. Their combined mass (water equivalent) is

$$ m_{\text{gain}} = 180 \text{ g (water)} + 20 \text{ g (equivalent of calorimeter)} = 200 \text{ g}. $$

The temperature rise of this 200 g is from $$25^\circ\text{C}$$ to $$31^\circ\text{C}$$, i.e. $$\Delta T = 31 - 25 = 6^\circ\text{C}$$. Using the specific heat of water $$c = 1 \text{ cal g}^{-1}{}^\circ\text{C}^{-1}$$,

$$ Q_{\text{gained}} = m_{\text{gain}} \, c \, \Delta T = 200 \times 1 \times 6 = 1200 \text{ cal}. $$

Heat lost is by the incoming steam. Two separate processes occur:

(i) Condensation: Each gram of steam releases its latent heat $$L = 540 \text{ cal g}^{-1}$$ when it turns into water at $$100^\circ\text{C}$$. Thus the heat released in condensation is $$m \times 540 \text{ cal}$$.

(ii) Cooling: The condensed water, whose mass is still $$m$$ grams, then cools from $$100^\circ\text{C}$$ to the final $$31^\circ\text{C}$$. The temperature drop is $$100 - 31 = 69^\circ\text{C}$$, so the heat released in cooling is

$$ Q_{\text{cool}} = m \, c \, \Delta T = m \times 1 \times 69 = 69m \text{ cal}. $$

Hence the total heat lost by the steam is

$$ Q_{\text{lost}} = m \times 540 + 69m = m(540 + 69) = m \times 609 \text{ cal}. $$

Applying conservation of energy,

$$ Q_{\text{lost}} = Q_{\text{gained}} \quad\Longrightarrow\quad m \times 609 = 1200. $$

Solving for $$m$$, we divide both sides by $$609$$:

$$ m = \frac{1200}{609}. $$

Carrying out the division,

$$ m \approx 1.972 \text{ g}. $$

The options given are 2 g, 4 g, 3.2 g, and 2.6 g. The value we obtained, $$1.972 \text{ g}$$, is closest to $$2 \text{ g}$$.

Hence, the correct answer is Option A.