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 finding the final temperature of the mixture when 300 grams of water at 25°C is added to 100 grams of ice at 0°C, we will follow these steps:

### Step 1: Identify the Heat Transfer
When ice at 0°C is added to water at 25°C, the heat from the water will be used to melt the ice. The water will cool down, while the ice will absorb heat and turn into water at 0°C.

### Step 2: Calculate the Heat Required to Melt the Ice
The heat required to melt the ice can be calculated using the formula:
\[ Q = m \cdot L_f \]
Where:
- \( Q \) = heat absorbed (in joules)
- \( m \) = mass of ice (in grams)
- \( L_f \) = latent heat of fusion of ice (approximately 80 cal/g or 334,000 J/kg)

Given:
- Mass of ice, \( m = 100 \) grams
- Latent heat of fusion, \( L_f = 80 \) cal/g = \( 80 \times 4.18 \) J/g = 334 J/g

Calculating heat required:
\[ Q = 100 \, \text{g} \times 334 \, \text{J/g} = 33400 \, \text{J} \]

### Step 3: Calculate the Heat Lost by the Water
The heat lost by the water as it cools from 25°C to 0°C can be calculated using:
\[ Q = m \cdot c \cdot \Delta T \]
Where:
- \( m \) = mass of water (in grams)
- \( c \) = specific heat capacity of water (approximately 4.18 J/g°C)
- \( \Delta T \) = change in temperature (in °C)

Given:
- Mass of water, \( m = 300 \) grams
- Initial temperature of water = 25°C
- Final temperature of water = 0°C
- Change in temperature, \( \Delta T = 25 - 0 = 25 \)°C

Calculating heat lost:
\[ Q = 300 \, \text{g} \times 4.18 \, \text{J/g°C} \times 25 \, \text{°C} = 31350 \, \text{J} \]

### Step 4: Compare Heat Gained and Lost
Now, we compare the heat gained by the ice and the heat lost by the water:
- Heat gained by ice = 33400 J
- Heat lost by water = 31350 J

Since the heat gained by the ice (33400 J) is greater than the heat lost by the water (31350 J), not all the ice will melt.

### Step 5: Calculate the Amount of Ice that Melts
The amount of ice that can melt using the heat lost by the water:
Using the heat lost by the water:
\[ Q = m \cdot L_f \]
Rearranging gives:
\[ m = \frac{Q}{L_f} \]
Where \( Q = 31350 \, \text{J} \) and \( L_f = 334 \, \text{J/g} \):
\[ m = \frac{31350 \, \text{J}}{334 \, \text{J/g}} \approx 93.75 \, \text{g} \]

### Step 6: Determine the Remaining Ice
The initial mass of ice was 100 grams, and after melting 93.75 grams, the remaining ice is:
\[ \text{Remaining ice} = 100 \, \text{g} - 93.75 \, \text{g} = 6.25 \, \text{g} \]

### Step 7: Final Temperature of the Mixture
Since there is still ice remaining and the system has reached thermal equilibrium, the final temperature of the mixture will be 0°C (as long as there is ice present, the temperature cannot rise above 0°C).

### Final Answer
The final temperature of the mixture is **0°C**.

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