To solve the problem of determining the amount of ice and water present at equilibrium when 10 grams of ice at -20°C is added to 10 grams of water at 50°C, we can follow these steps:
### Step 1: Calculate the heat required to warm the ice to 0°C
The formula to calculate the heat (Q) required to change the temperature of a substance is:
\[ Q = m \cdot S \cdot \Delta T \]
Where:
- \( m \) = mass of the ice = 10 grams
- \( S \) = specific heat of ice = 0.5 cal/g°C
- \( \Delta T \) = change in temperature = 0°C - (-20°C) = 20°C
Calculating:
\[ Q_1 = 10 \, \text{g} \cdot 0.5 \, \text{cal/g°C} \cdot 20 \, \text{°C} = 100 \, \text{cal} \]
### Step 2: Calculate the heat released by the water when it cools to 0°C
Using the same formula for the water:
- \( m \) = mass of the water = 10 grams
- \( S \) = specific heat of water = 1 cal/g°C
- \( \Delta T \) = change in temperature = 0°C - 50°C = -50°C
Calculating:
\[ Q_2 = 10 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot (-50 \, \text{°C}) = -500 \, \text{cal} \]
(Note: The negative sign indicates heat is released.)
### Step 3: Determine the net heat available for melting the ice
The net heat available after the ice has warmed up to 0°C and the water has cooled down to 0°C can be calculated as:
\[ Q_{\text{net}} = Q_2 - Q_1 \]
\[ Q_{\text{net}} = -500 \, \text{cal} - 100 \, \text{cal} = -400 \, \text{cal} \]
Since we are interested in the absolute value of heat available for melting ice:
\[ Q_{\text{available}} = 400 \, \text{cal} \]
### Step 4: Calculate how much ice can melt with the available heat
The latent heat of fusion of ice is approximately 80 cal/g. We can find out how much ice can melt using the formula:
\[ \text{mass of ice melted} = \frac{Q_{\text{available}}}{\text{latent heat of fusion}} \]
Calculating:
\[ \text{mass of ice melted} = \frac{400 \, \text{cal}}{80 \, \text{cal/g}} = 5 \, \text{g} \]
### Step 5: Determine the final amounts of ice and water
Initially, we had:
- Ice: 10 g
- Water: 10 g
After melting 5 g of ice:
- Remaining ice = \( 10 \, \text{g} - 5 \, \text{g} = 5 \, \text{g} \)
- Total water = \( 10 \, \text{g} + 5 \, \text{g} = 15 \, \text{g} \)
### Final Answer
At equilibrium, there will be **5 grams of ice** and **15 grams of water** present.
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