Ice at `-20^(@)C` mixed with 200g water at `25^(@)C`. If temperature of mixture is `10^(@)C` then mass of ice is -

Options

A. 30 gm
B. 20 gm
C. 15 gm
D. 40 gm

Answer

To solve the problem of finding the mass of ice at -20°C mixed with 200g of water at 25°C, which results in a final temperature of 10°C, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Variables**:
- Let the mass of ice be \( m \) grams.
- The mass of water is given as \( 200 \) grams.
- The specific heat of ice is \( c_{ice} = 2.1 \, \text{J/g°C} \).
- The specific heat of water is \( c_{water} = 4.18 \, \text{J/g°C} \).
- The latent heat of fusion of ice is \( L_f = 334 \, \text{J/g} \).

2. **Calculate Heat Gained by Ice**:
- The ice first warms up from -20°C to 0°C:
\[ q_1 = m \cdot c_{ice} \cdot (0 - (-20)) = m \cdot 2.1 \cdot 20 = 42m \, \text{J} \]
- Then, the ice melts at 0°C:
\[ q_2 = m \cdot L_f = m \cdot 334 \, \text{J} \]
- After melting, the resulting water warms up from 0°C to 10°C:
\[ q_3 = m \cdot c_{water} \cdot (10 - 0) = m \cdot 4.18 \cdot 10 = 41.8m \, \text{J} \]

3. **Calculate Heat Lost by Water**:
- The water cools down from 25°C to 10°C:
\[ q_4 = 200 \cdot c_{water} \cdot (10 - 25) = 200 \cdot 4.18 \cdot (-15) = -12540 \, \text{J} \]

4. **Set Up the Heat Balance Equation**:
- The heat gained by the ice (from steps 2) equals the heat lost by the water (from step 3):
\[ q_1 + q_2 + q_3 = -q_4 \]
\[ 42m + 334m + 41.8m = 12540 \]
\[ 417.8m = 12540 \]

5. **Solve for \( m \)**:
\[ m = \frac{12540}{417.8} \approx 30 \, \text{grams} \]

### Final Answer:
The mass of ice is approximately **30 grams**.