300 grams of water at `25^@ C` is added to 100 grams of ice at `0^@ C.` The final temperature of the mixture is _____ `^@C`

Explanation

To solve the problem of mixing 300 grams of water at 25°C with 100 grams of ice at 0°C, we need to follow these steps:

### Step 1: Calculate the heat released by the water
The heat released by the water as it cools from 25°C to 0°C can be calculated using the formula:
\[ Q_{\text{released}} = m \cdot s \cdot \Delta T \]

Where:
- \( m = 300 \, \text{g} \) (mass of water)
- \( s = 1 \, \text{cal/g°C} \) (specific heat of water)
- \( \Delta T = 25°C - 0°C = 25°C \) (temperature change)

Converting calories to joules:
\[ 1 \, \text{cal} = 4.2 \, \text{J} \]
So,
\[ Q_{\text{released}} = 300 \cdot 1 \cdot 25 \cdot 4.2 = 31500 \, \text{J} \]

### Step 2: Calculate the heat absorbed by the ice
The ice will first absorb heat to melt into water at 0°C, and then the resulting water will absorb heat to increase its temperature to the final temperature (which we will denote as \( T_f \)).

1. **Heat absorbed to melt the ice:**
\[ Q_{\text{melt}} = m \cdot L \]
Where:
- \( m = 100 \, \text{g} \) (mass of ice)
- \( L = 80 \, \text{cal/g} \) (latent heat of fusion)

Converting to joules:
\[ Q_{\text{melt}} = 100 \cdot 80 \cdot 4.2 = 33600 \, \text{J} \]

2. **Heat absorbed to raise the temperature of melted ice (water at 0°C) to \( T_f \):**
\[ Q_{\text{heat}} = m \cdot s \cdot \Delta T \]
Where:
- \( m = 100 \, \text{g} \) (mass of melted ice)
- \( s = 1 \, \text{cal/g°C} \)
- \( \Delta T = T_f - 0°C \)

Thus,
\[ Q_{\text{heat}} = 100 \cdot 1 \cdot (T_f - 0) \cdot 4.2 = 420 \cdot T_f \]

### Step 3: Set up the energy balance equation
The heat lost by the water equals the heat gained by the ice:
\[ Q_{\text{released}} = Q_{\text{melt}} + Q_{\text{heat}} \]
Substituting the values we calculated:
\[ 31500 = 33600 + 420 \cdot T_f \]

### Step 4: Solve for \( T_f \)
Rearranging the equation:
\[ 420 \cdot T_f = 31500 - 33600 \]
\[ 420 \cdot T_f = -2100 \]
\[ T_f = \frac{-2100}{420} = -5°C \]

Since a negative temperature does not make sense in this context, it indicates that not all the ice has melted.

### Step 5: Calculate the amount of ice that remains
The heat absorbed by the ice is greater than the heat released by the water, meaning not all ice can melt. The amount of heat required to melt all the ice is:
\[ Q_{\text{total absorbed}} = Q_{\text{melt}} + Q_{\text{heat}} \]
The heat deficit is:
\[ Q_{\text{deficit}} = Q_{\text{melt}} - Q_{\text{released}} = 33600 - 31500 = 2100 \, \text{J} \]

### Step 6: Calculate the mass of ice that remains
Using the latent heat of fusion:
\[ Q_{\text{deficit}} = m \cdot L \]
\[ 2100 = m \cdot 80 \cdot 4.2 \]
\[ m = \frac{2100}{336} = 6.25 \, \text{g} \]

### Step 7: Calculate the final amount of water
The amount of ice that melted is:
\[ 100 - 6.25 = 93.75 \, \text{g} \]
The total amount of water at 0°C is:
\[ 300 \, \text{g} + 93.75 \, \text{g} = 393.75 \, \text{g} \]

### Final Answer
The final temperature of the mixture is 0°C, with 6.25 grams of ice remaining unmelted.

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