A lump of ice of 0.1 kg at `-10^(@)C` is put in 0.15 kg of water at `20^(@)C`. How much water and ice will be found in the mixture whenit has reached thermal equilibrium? Specific heat of ice `=0.5kcal kg^(-1).^(@)C^(-1)` and its latent heat of melting `=80` kcal `kg^(-1)`

Answer

To solve the problem, we need to analyze the heat transfer between the ice and the water until they reach thermal equilibrium. Here’s a step-by-step solution:

### Step 1: Calculate the heat required to warm the ice from -10°C to 0°C.
The formula to calculate the heat required to change the temperature of a substance is:
\[ Q = m \cdot c \cdot \Delta T \]
Where:
- \( Q \) = heat absorbed (in kcal)
- \( m \) = mass of the substance (in kg)
- \( c \) = specific heat capacity (in kcal/kg°C)
- \( \Delta T \) = change in temperature (in °C)

For the ice:
- Mass of ice, \( m = 0.1 \, \text{kg} \)
- Specific heat of ice, \( c = 0.5 \, \text{kcal/kg°C} \)
- Change in temperature, \( \Delta T = 0 - (-10) = 10 \, \text{°C} \)

Calculating the heat required:
\[ Q_{\text{ice}} = 0.1 \, \text{kg} \cdot 0.5 \, \text{kcal/kg°C} \cdot 10 \, \text{°C} = 0.5 \, \text{kcal} \]

### Step 2: Calculate the heat released by the water as it cools from 20°C to 0°C.
For the water:
- Mass of water, \( m = 0.15 \, \text{kg} \)
- Specific heat of water, \( c = 1 \, \text{kcal/kg°C} \)
- Change in temperature, \( \Delta T = 20 - 0 = 20 \, \text{°C} \)

Calculating the heat released:
\[ Q_{\text{water}} = 0.15 \, \text{kg} \cdot 1 \, \text{kcal/kg°C} \cdot 20 \, \text{°C} = 3 \, \text{kcal} \]

### Step 3: Determine the heat balance.
The heat released by the water (3 kcal) is greater than the heat required to warm the ice (0.5 kcal). The remaining heat after warming the ice is:
\[ Q_{\text{remaining}} = Q_{\text{water}} - Q_{\text{ice}} = 3 \, \text{kcal} - 0.5 \, \text{kcal} = 2.5 \, \text{kcal} \]

### Step 4: Use the remaining heat to melt the ice.
The latent heat of fusion of ice is given as 80 kcal/kg. To find out how much ice can be melted with the remaining heat:
Let \( m \) be the mass of ice melted.
\[ Q_{\text{remaining}} = m \cdot L \]
Where \( L = 80 \, \text{kcal/kg} \).

Setting up the equation:
\[ 2.5 \, \text{kcal} = m \cdot 80 \, \text{kcal/kg} \]
Solving for \( m \):
\[ m = \frac{2.5 \, \text{kcal}}{80 \, \text{kcal/kg}} = 0.03125 \, \text{kg} \]

### Step 5: Calculate the final amounts of water and ice.
Initially, we had:
- Ice: 0.1 kg
- Water: 0.15 kg

After melting, the amount of ice remaining:
\[ \text{Ice remaining} = 0.1 \, \text{kg} - 0.03125 \, \text{kg} = 0.06875 \, \text{kg} \]

The total amount of water after melting:
\[ \text{Total water} = 0.15 \, \text{kg} + 0.03125 \, \text{kg} = 0.18125 \, \text{kg} \]

### Final Answer:
- Amount of water: 0.18125 kg
- Amount of ice: 0.06875 kg