How much ice must be added to 100 g water at `30^@C` in order to reduce its temperature to `20^@C`?

Options

A. 10 g
B. 80 g
C. 400 g
D. None of these

Explanation

To solve the problem of how much ice must be added to 100 g of water at 30°C to reduce its temperature to 20°C, we can follow these steps:

### Step 1: Understand the Heat Exchange
When ice is added to warm water, the warm water loses heat and cools down, while the ice gains heat and melts. The heat lost by the water will equal the heat gained by the ice.

### Step 2: Write the Heat Loss and Gain Equations
1. **Heat lost by water (Q_water)**:
\[ Q_{\text{water}} = m_w \cdot c_w \cdot \Delta T_w \]
where:
- \( m_w = 100 \, \text{g} \) (mass of water)
- \( c_w = 1 \, \text{cal/g°C} \) (specific heat of water)
- \( \Delta T_w = (30 - 20) \, \text{°C} = 10 \, \text{°C} \)

So,
\[ Q_{\text{water}} = 100 \cdot 1 \cdot 10 = 1000 \, \text{cal} \]

2. **Heat gained by ice (Q_ice)**:
The ice first melts and then warms up to the final temperature of the water.
\[ Q_{\text{ice}} = m_i \cdot L_f + m_i \cdot c_i \cdot \Delta T_i \]
where:
- \( m_i \) = mass of ice (unknown, let it be \( x \))
- \( L_f = 80 \, \text{cal/g} \) (latent heat of fusion of ice)
- \( c_i = 1 \, \text{cal/g°C} \) (specific heat of water)
- \( \Delta T_i = (20 - 0) \, \text{°C} = 20 \, \text{°C} \)

So,
\[ Q_{\text{ice}} = x \cdot 80 + x \cdot 1 \cdot 20 = 80x + 20x = 100x \, \text{cal} \]

### Step 3: Set the Heat Lost Equal to Heat Gained
Since the heat lost by the water equals the heat gained by the ice:
\[ Q_{\text{water}} = Q_{\text{ice}} \]
\[ 1000 = 100x \]

### Step 4: Solve for \( x \)
Now, we can solve for \( x \):
\[ x = \frac{1000}{100} = 10 \, \text{g} \]

### Conclusion
Thus, the amount of ice that must be added to 100 g of water at 30°C to reduce its temperature to 20°C is **10 grams**.

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