A tap supplies water at `10^@ C` and another tap at `100^@ C`. How much hot water must be taken so that we get `20 kg` of water at `35^@ C`

Options

A. `40//9 kg`
B. `50//9 kg`
C. `20//9 kg`
D. `130//9 kg`

Explanation

To solve the problem of how much hot water must be taken to obtain 20 kg of water at 35°C, we can use the principle of calorimetry, which states that the heat lost by the hot water will equal the heat gained by the cold water.

### Step-by-Step Solution:

1. **Define Variables:**
- Let \( m \) be the mass of hot water taken at 100°C.
- The mass of cold water taken at 10°C will then be \( 20 - m \) kg, since the total mass of water is 20 kg.

2. **Set Up the Heat Equation:**
- The heat lost by the hot water can be expressed as:
\[ Q_{\text{hot}} = m \cdot c \cdot (T_{\text{initial hot}} - T_{\text{final}}) \]
where \( c \) is the specific heat capacity of water (which is 1 cal/g°C), \( T_{\text{initial hot}} = 100°C \), and \( T_{\text{final}} = 35°C \).
- Therefore, the heat lost by the hot water is:
\[ Q_{\text{hot}} = m \cdot 1 \cdot (100 - 35) = m \cdot 65 \]

3. **Calculate Heat Gained by Cold Water:**
- The heat gained by the cold water can be expressed as:
\[ Q_{\text{cold}} = (20 - m) \cdot c \cdot (T_{\text{final}} - T_{\text{initial cold}}) \]
where \( T_{\text{initial cold}} = 10°C \).
- Therefore, the heat gained by the cold water is:
\[ Q_{\text{cold}} = (20 - m) \cdot 1 \cdot (35 - 10) = (20 - m) \cdot 25 \]

4. **Set Heat Lost Equal to Heat Gained:**
- According to the principle of calorimetry:
\[ Q_{\text{hot}} = Q_{\text{cold}} \]
- Substituting the expressions we derived:
\[ m \cdot 65 = (20 - m) \cdot 25 \]

5. **Solve for \( m \):**
- Expanding the right side:
\[ m \cdot 65 = 500 - 25m \]
- Rearranging gives:
\[ 65m + 25m = 500 \]
\[ 90m = 500 \]
\[ m = \frac{500}{90} = \frac{50}{9} \text{ kg} \]

6. **Conclusion:**
- Therefore, the mass of hot water needed is:
\[ m \approx 5.56 \text{ kg} \]

### Final Answer:
You must take approximately \( \frac{50}{9} \) kg (or about 5.56 kg) of hot water at 100°C to obtain 20 kg of water at 35°C.