5 g of water at `30^@C` and 5 g of ice at `-20^@C` are mixed together in a calorimeter Find the final temperature of the mixure. Assume water equivalent of calorimeter to be negligible, sp. Heats of ice and water are 0.5 and `1 cal//g C^@`, and latent heat of ice is `80 cal//g`

Options

A. `0^(@)C`
B. `10^(@)C`
C. `-30^(@)C`
D. `gt10^(@)C`

Explanation

To solve the problem of finding the final temperature when mixing 5 g of water at 30°C with 5 g of ice at -20°C, we can follow these steps:

### Step 1: Calculate the heat required to warm the ice from -20°C to 0°C

The formula for calculating the heat (Q) required to change the temperature of a substance is:

\[ Q = m \cdot c \cdot \Delta T \]

Where:
- \( m \) = mass of the substance (in grams)
- \( c \) = specific heat capacity (in cal/g°C)
- \( \Delta T \) = change in temperature (in °C)

For the ice:
- Mass of ice, \( m = 5 \, \text{g} \)
- Specific heat of ice, \( c = 0.5 \, \text{cal/g°C} \)
- Change in temperature, \( \Delta T = 0 - (-20) = 20 \, \text{°C} \)

Now, substituting the values:

\[ Q_1 = 5 \, \text{g} \cdot 0.5 \, \text{cal/g°C} \cdot 20 \, \text{°C} = 50 \, \text{cal} \]

### Step 2: Calculate the heat required to melt the ice at 0°C

The formula for calculating the heat required to melt ice is:

\[ Q = m \cdot L \]

Where:
- \( L \) = latent heat of fusion (in cal/g)

For the ice:
- Mass of ice, \( m = 5 \, \text{g} \)
- Latent heat of fusion, \( L = 80 \, \text{cal/g} \)

Now, substituting the values:

\[ Q_2 = 5 \, \text{g} \cdot 80 \, \text{cal/g} = 400 \, \text{cal} \]

### Step 3: Calculate the total heat required to convert ice at -20°C to water at 0°C

The total heat required (Q_total) to convert the ice at -20°C to water at 0°C is the sum of the heat required to warm the ice and the heat required to melt it:

\[ Q_{\text{total}} = Q_1 + Q_2 = 50 \, \text{cal} + 400 \, \text{cal} = 450 \, \text{cal} \]

### Step 4: Calculate the heat released by the water as it cools from 30°C to 0°C

Using the same formula as before for the water:

\[ Q = m \cdot c \cdot \Delta T \]

For the water:
- Mass of water, \( m = 5 \, \text{g} \)
- Specific heat of water, \( c = 1 \, \text{cal/g°C} \)
- Change in temperature, \( \Delta T = 30 - 0 = 30 \, \text{°C} \)

Now, substituting the values:

\[ Q_{\text{water}} = 5 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 30 \, \text{°C} = 150 \, \text{cal} \]

### Step 5: Compare heat gained and heat lost

We have:
- Heat required to convert ice to water: \( 450 \, \text{cal} \)
- Heat released by water: \( 150 \, \text{cal} \)

Since the heat released by the water (150 cal) is less than the heat required to convert the ice (450 cal), the ice will not completely melt.

### Final Temperature

The ice will warm up to 0°C, and some of it will melt, but not all. Therefore, the final temperature of the mixture will be:

\[ \text{Final Temperature} = 0 \, \text{°C} \]