To solve the problem of mixing ice and water, we will follow these steps:
### Step 1: Identify the given values
- Mass of ice, \( m_i = 2 \, \text{kg} = 2000 \, \text{g} \)
- Initial temperature of ice, \( T_i = -20^\circ C \)
- Mass of water, \( m_w = 5 \, \text{kg} = 5000 \, \text{g} \)
- Initial temperature of water, \( T_w = 20^\circ C \)
- Specific heat of ice, \( s_i = 0.5 \, \text{cal/g}^\circ C \)
- Specific heat of water, \( s_w = 1 \, \text{cal/g}^\circ C \)
- Latent heat of fusion of ice, \( L = 80 \, \text{cal/g} \)
### Step 2: Calculate the heat gained by the ice to reach 0°C
The heat gained by the ice to raise its temperature from \(-20^\circ C\) to \(0^\circ C\) is given by:
\[
Q_1 = m_i \cdot s_i \cdot \Delta T
\]
where \(\Delta T = 0 - (-20) = 20^\circ C\).
Substituting the values:
\[
Q_1 = 2000 \, \text{g} \cdot 0.5 \, \text{cal/g}^\circ C \cdot 20^\circ C = 20000 \, \text{cal}
\]
### Step 3: Calculate the heat lost by the water as it cools to 0°C
The heat lost by the water to cool from \(20^\circ C\) to \(0^\circ C\) is given by:
\[
Q_2 = m_w \cdot s_w \cdot \Delta T
\]
where \(\Delta T = 20 - 0 = 20^\circ C\).
Substituting the values:
\[
Q_2 = 5000 \, \text{g} \cdot 1 \, \text{cal/g}^\circ C \cdot 20^\circ C = 100000 \, \text{cal}
\]
### Step 4: Determine the heat required to melt the ice
The heat required to melt the ice at \(0^\circ C\) is given by:
\[
Q_3 = m_f \cdot L
\]
where \(m_f\) is the mass of ice that will melt.
### Step 5: Set up the energy balance equation
The heat lost by the water will be equal to the heat gained by the ice plus the heat required to melt the ice:
\[
Q_2 = Q_1 + Q_3
\]
Substituting the known values:
\[
100000 \, \text{cal} = 20000 \, \text{cal} + m_f \cdot 80 \, \text{cal/g}
\]
### Step 6: Solve for \(m_f\)
Rearranging the equation:
\[
100000 - 20000 = m_f \cdot 80
\]
\[
80000 = m_f \cdot 80
\]
\[
m_f = \frac{80000}{80} = 1000 \, \text{g} = 1 \, \text{kg}
\]
### Step 7: Calculate the final amount of water in the mixture
The final amount of water in the mixture will be the mass of the original water plus the mass of the melted ice:
\[
\text{Final mass of water} = m_w + m_f = 5000 \, \text{g} + 1000 \, \text{g} = 6000 \, \text{g} = 6 \, \text{kg}
\]
### Final Answer
The final amount of water in the mixture will be **6 kg**.
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