According to the principle of calorimetry, Heat lost = Heat gained. Let $m$ be the mass of steam condensed. The heat lost by $m$ gram of steam at $100^{\circ}\text{C}$ to condense into water at $100^{\circ}\text{C}$ and then cool down to $80^{\circ}\text{C}$ is: $Q_{\text{lost}} = m \times L_v + m \times s \times \Delta T_{\text{steam}}$ $Q_{\text{lost}} = m \times 540 + m \times 1 \times (100 - 80)$ $Q_{\text{lost}} = 540m + 20m = 560m$

The heat gained by $20\text{ g}$ of water at $10^{\circ}\text{C}$ to reach $80^{\circ}\text{C}$ is: $Q_{\text{gained}} = M \times s \times \Delta T_{\text{water}}$ $Q_{\text{gained}} = 20 \times 1 \times (80 - 10) = 20 \times 70 = 1400\text{ cal}$

Equating the heat lost and heat gained: $560m = 1400$ $m = \frac{1400}{560} = 2.5\text{ g}$

Thus, the total mass of water present at the end is the initial mass of water plus the mass of condensed steam: Total mass of water $= 20\text{ g} + 2.5\text{ g} = 22.5\text{ g}$.