Suppose that 40 g of solid mercury at its freezing point `(-39^(@)C)` is dropped into a mixture of water and ice of `0^(@)C`. After equilibrium is achieved, the mercury -ice -water mixture is still at `0^(@)C`. How much additiional ice is produced by the addition of the mercury?

Answer

To solve the problem of how much additional ice is produced by the addition of 40 g of solid mercury at its freezing point (-39°C) into a mixture of water and ice at 0°C, we will follow these steps:

### Step 1: Calculate the heat required to convert solid mercury to liquid mercury at 0°C.
The latent heat of fusion of mercury is given as 2.295 kJ/mol. First, we need to convert the mass of mercury into moles.

1. **Molar mass of mercury (Hg)** = 200 g/mol.
2. **Mass of mercury** = 40 g.

\[ \text{Number of moles of mercury} = \frac{\text{mass}}{\text{molar mass}} = \frac{40 \, \text{g}}{200 \, \text{g/mol}} = 0.2 \, \text{mol} \]

Now, calculate the heat required (Q1) to convert solid mercury at -39°C to liquid mercury at 0°C using the formula:

\[ Q_1 = n \times L_f \]

Where:
- \( n \) = number of moles of mercury = 0.2 mol
- \( L_f \) = latent heat of fusion of mercury = 2.295 kJ/mol = 2295 J/mol

\[ Q_1 = 0.2 \, \text{mol} \times 2295 \, \text{J/mol} = 459 \, \text{J} \]

### Step 2: Calculate the heat required to raise the temperature of the liquid mercury from -39°C to 0°C.
The specific heat of liquid mercury is given as 336 J/kg°C.

1. Convert the mass of mercury to kg:
\[ \text{Mass of mercury} = 40 \, \text{g} = 0.04 \, \text{kg} \]

2. Calculate the heat required (Q2) to raise the temperature of mercury from -39°C to 0°C:

\[ Q_2 = m \times c \times \Delta T \]

Where:
- \( m \) = mass of mercury = 0.04 kg
- \( c \) = specific heat of mercury = 336 J/kg°C
- \( \Delta T = 0 - (-39) = 39°C\)

\[ Q_2 = 0.04 \, \text{kg} \times 336 \, \text{J/kg°C} \times 39 \, \text{°C} = 526.08 \, \text{J} \]

### Step 3: Calculate the total heat absorbed by mercury.
The total heat absorbed (Q_total) by mercury is:

\[ Q_{\text{total}} = Q_1 + Q_2 = 459 \, \text{J} + 526.08 \, \text{J} = 985.08 \, \text{J} \]

### Step 4: Calculate how much ice is produced from the heat lost by the water.
The heat lost by the water (which turns into ice) can be calculated using the latent heat of fusion of ice, which is approximately 334,000 J/kg.

Let \( m \) be the mass of water that turns into ice:

\[ Q_{\text{lost}} = m \times L_f \]

Where \( L_f \) for ice = 334,000 J/kg.

Setting the heat lost equal to the heat gained by mercury:

\[ 985.08 \, \text{J} = m \times 334,000 \, \text{J/kg} \]

Solving for \( m \):

\[ m = \frac{985.08 \, \text{J}}{334,000 \, \text{J/kg}} \approx 0.00295 \, \text{kg} = 2.95 \, \text{g} \]

### Conclusion
Thus, the additional ice produced by the addition of the mercury is approximately **2.95 g**.

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