A mass of 1 kg of ice at 0^ ^ ^ C is mixed with 1 kg of steam at...

Heat required to convert $1 \text{ kg}$ of ice at 0^{}^{{}^{\circ}} \text{C} to water at 100^{}^{{}^{\circ}} \text{C} is given by:

$Q_1 = m_{\text{ice}} L_f + m_{\text{ice}} C_w \Delta T$

$Q_1 = 1 \times 3.36 \times 10^5 + 1 \times 4200 \times 100$

$Q_1 = 3.36 \times 10^5 + 4.20 \times 10^5 = 7.56 \times 10^5 \text{ J}$

Heat released if the entire $1 \text{ kg}$ of steam at 100^{}^{{}^{\circ}} \text{C} condenses to water at 100^{}^{{}^{\circ}} \text{C}:

$Q_2 = m_{\text{steam}} L_v = 1 \times 2.26 \times 10^6 = 22.6 \times 10^5 \text{ J}$

Since $Q_2 > Q_1$, the available heat from steam is more than sufficient to melt the ice and raise its temperature to 100^{}^{{}^{\circ}} \text{C}. Thus, the final temperature of the mixture will be 100^{}^{{}^{\circ}} \text{C} and only a fraction of the steam will condense.

Let $m$ be the mass of steam that condenses to provide the required heat $Q_1$:

$m L_v = Q_1$

$m \times 2.26 \times 10^6 = 7.56 \times 10^5$

$m = \dfrac{7.56 \times 10^5}{2.26 \times 10^6} \approx 0.3345 \text{ kg} = 334.5 \text{ g}$

The mass of steam remaining in the mixture:

$m_{\text{remaining steam}} = 1000 \text{ g} - 334.5 \text{ g} = 665.5 \text{ g} \approx 665 \text{ g}$

The total mass of water in the mixture is the initial mass of ice plus the mass of condensed steam:

$m_{\text{water}} = 1000 \text{ g} + 334.5 \text{ g} = 1334.5 \text{ g} \approx 1.335 \text{ kg}$

The final composition of the system is $665 \text{ g}$ of steam and $1.335 \text{ kg}$ of water.