100 g of ice at `0^(@)` ismixed with 100 g of water at `100^(@)C`.What will be the final temperature of the mixture? (Assume no heat loss)

Options

A. `20^(@)C`
B. `40^(@)C`
C. `30^(@)C`
D. `10^(@)C`

Explanation

To solve the problem of finding the final temperature of the mixture of ice and water, we can follow these steps:

### Step 1: Identify the heat exchange components
We have:
- Mass of ice (m_ice) = 100 g
- Initial temperature of ice (T_initial_ice) = 0 °C
- Mass of water (m_water) = 100 g
- Initial temperature of water (T_initial_water) = 100 °C

### Step 2: Calculate the heat required to melt the ice
The heat required to melt the ice can be calculated using the formula:
\[ Q_{melt} = m_{ice} \times L_f \]
where \( L_f \) is the latent heat of fusion of ice, which is approximately 334 J/g.

So,
\[ Q_{melt} = 100 \, \text{g} \times 334 \, \text{J/g} = 33400 \, \text{J} \]

### Step 3: Calculate the heat lost by the water as it cools down
The heat lost by the water as it cools from 100 °C to the final temperature (T_f) can be calculated using the formula:
\[ Q_{lost} = m_{water} \times c_{water} \times (T_{initial\_water} - T_f) \]
where \( c_{water} \) is the specific heat capacity of water, approximately 4.18 J/g°C.

So,
\[ Q_{lost} = 100 \, \text{g} \times 4.18 \, \text{J/g°C} \times (100 - T_f) \]

### Step 4: Set up the heat balance equation
Since there is no heat loss to the surroundings, the heat gained by the ice will equal the heat lost by the water:
\[ Q_{melt} + Q_{gain} = Q_{lost} \]

Since the ice will absorb heat to melt and then potentially to raise its temperature to T_f, we can express the heat gained by the melted ice as:
\[ Q_{gain} = m_{ice} \times c_{water} \times (T_f - 0) = 100 \, \text{g} \times 4.18 \, \text{J/g°C} \times T_f \]

Thus, the equation becomes:
\[ 33400 + 100 \times 4.18 \times T_f = 100 \times 4.18 \times (100 - T_f) \]

### Step 5: Solve for T_f
Expanding and rearranging the equation:
\[ 33400 + 418T_f = 41800 - 418T_f \]
\[ 33400 + 418T_f + 418T_f = 41800 \]
\[ 33400 + 836T_f = 41800 \]
\[ 836T_f = 41800 - 33400 \]
\[ 836T_f = 8400 \]
\[ T_f = \frac{8400}{836} \approx 10.05 °C \]

### Final Answer
The final temperature of the mixture is approximately **10.05 °C**.

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