Answer the following in English showing all steps used

The heat released by the steam as it condenses into water at $100 \text{ °C}$ is given by the formula $Q = m \cdot L_v$, where $m$ is the mass of the steam and $L_v$ is the latent heat of vaporization of water ($2260 \text{ J/g}$).

$$Q_{steam} = m_{steam} \cdot L_v$$ $$Q_{steam} = 75 \text{ g} \times 2260 \text{ J/g}$$ $$Q_{steam} = 169500 \text{ J}$$

The problem states that the heat released by the steam is just sufficient to melt the ice. The heat absorbed by the ice as it melts at $0 \text{ °C}$ is given by $Q = m \cdot L_f$, where $m$ is the mass of the ice and $L_f$ is the latent heat of fusion of ice ($334 \text{ J/g}$).

$$Q_{ice} = m_{ice} \cdot L_f$$

Since $Q_{steam} = Q_{ice}$:

$$169500 \text{ J} = m_{ice} \cdot 334 \text{ J/g}$$ $$m_{ice} = \frac{169500 \text{ J}}{334 \text{ J/g}}$$ $$m_{ice} \approx 507.485 \text{ g}$$

Rounding to three significant figures:

$$m_{ice} \approx 507 \text{ g}$$

The mass of the ice is $\boxed{\text{507 g}}$.

After the ice has completely melted, the system contains:

• Water from condensed steam: $m_s = 75 \text{ g}$ at $100 \text{ °C}$.
• Water from melted ice: $m_{ice} \approx 507.485 \text{ g}$ at $0 \text{ °C}$. These two quantities of water will mix and reach an equilibrium temperature, $T_f$. The specific heat capacity of water ($c_w$) is $4.18 \text{ J/(g} \cdot \text{°C)}$.

In an insulated calorimeter, the heat lost by the hotter substance equals the heat gained by the colder substance.

$$Q_{lost} = Q_{gained}$$ $$m_s \cdot c_w \cdot (T_{initial\_steam\_water} - T_f) = m_{ice} \cdot c_w \cdot (T_f - T_{initial\_melted\_ice\_water})$$ $$75 \text{ g} \cdot c_w \cdot (100 \text{ °C} - T_f) = 507.485 \text{ g} \cdot c_w \cdot (T_f - 0 \text{ °C})$$

The specific heat capacity of water ($c_w$) cancels out from both sides:

$$75 \cdot (100 - T_f) = 507.485 \cdot T_f$$

$$7500 - 75 T_f = 507.485 T_f$$ $$7500 = 507.485 T_f + 75 T_f$$ $$7500 = (507.485 + 75) T_f$$ $$7500 = 582.485 T_f$$ $$T_f = \frac{7500}{582.485}$$ $$T_f \approx 12.876 \text{ °C}$$

Rounding to three significant figures:

$$T_f \approx 12.9 \text{ °C}$$

The equilibrium temperature of the mixture is $\boxed{\text{12.9 °C}}$.