How many gm of ice at `-20^(@)C` are needed to cool 200 gm of water from `25^(@)C` to `10^(@)C`?

Options

A. 20 gm
B. 30 gm
C. 40 gm
D. 50 gm

Explanation

To solve the problem of how many grams of ice at -20°C are needed to cool 200 grams of water from 25°C to 10°C, we can follow these steps:

### Step 1: Calculate the heat lost by the water
The heat lost by the water can be calculated using the formula:
\[ Q = m \cdot c \cdot \Delta T \]
where:
- \( m \) = mass of water = 200 g
- \( c \) = specific heat capacity of water = 1 cal/g°C
- \( \Delta T \) = change in temperature = \( (25 - 10) \) °C = 15 °C

Substituting the values:
\[ Q = 200 \, \text{g} \cdot 1 \, \text{cal/g°C} \cdot 15 \, \text{°C} = 3000 \, \text{cal} \]

### Step 2: Calculate the heat gained by the ice
The heat gained by the ice involves three processes:
1. Heating the ice from -20°C to 0°C
2. Melting the ice at 0°C to water
3. Heating the resulting water from 0°C to 10°C

#### 2.1: Heat required to raise the temperature of ice from -20°C to 0°C
\[ Q_1 = m \cdot c_{ice} \cdot \Delta T \]
where:
- \( c_{ice} \) = specific heat capacity of ice = 0.5 cal/g°C
- \( \Delta T = (0 - (-20)) \) °C = 20 °C

So,
\[ Q_1 = m \cdot 0.5 \, \text{cal/g°C} \cdot 20 \, \text{°C} = 10m \, \text{cal} \]

#### 2.2: Heat required to melt the ice
\[ Q_2 = m \cdot L_f \]
where:
- \( L_f \) = latent heat of fusion of ice = 80 cal/g

Thus,
\[ Q_2 = m \cdot 80 \, \text{cal} \]

#### 2.3: Heat required to raise the temperature of water from 0°C to 10°C
\[ Q_3 = m \cdot c_{water} \cdot \Delta T \]
where:
- \( c_{water} \) = specific heat capacity of water = 1 cal/g°C
- \( \Delta T = (10 - 0) \) °C = 10 °C

So,
\[ Q_3 = m \cdot 1 \, \text{cal/g°C} \cdot 10 \, \text{°C} = 10m \, \text{cal} \]

### Step 3: Total heat gained by the ice
The total heat gained by the ice is:
\[ Q_{ice} = Q_1 + Q_2 + Q_3 = 10m + 80m + 10m = 100m \, \text{cal} \]

### Step 4: Equate the heat lost by the water to the heat gained by the ice
Setting the heat lost by the water equal to the heat gained by the ice:
\[ 3000 \, \text{cal} = 100m \]

### Step 5: Solve for \( m \)
\[ m = \frac{3000}{100} = 30 \, \text{g} \]

### Conclusion
Thus, **30 grams of ice at -20°C are needed to cool 200 grams of water from 25°C to 10°C**.

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