Calculate the resulting temperature when 20 g of boiling water is poured into an ice-cold brass vessel `("specific heat = 0.1 cal g"^(-1) .^(@)C^(-1))` of mass 100 g

Options

A. `66.66.^(@)C`
B. `6.66.^(@)C`
C. `0.66.^(@)C`
D. `50.^(@)C`

Answer

To solve the problem of calculating the resulting temperature when 20 g of boiling water is poured into an ice-cold brass vessel, we will use the principle of conservation of energy. The heat lost by the boiling water will be equal to the heat gained by the brass vessel.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of boiling water, \( m_w = 20 \, \text{g} \)
- Initial temperature of water, \( T_w = 100^\circ \text{C} \)
- Mass of brass vessel, \( m_v = 100 \, \text{g} \)
- Initial temperature of brass vessel, \( T_v = 0^\circ \text{C} \)
- Specific heat of water, \( c_w = 1 \, \text{cal/g}^\circ \text{C} \)
- Specific heat of brass, \( c_v = 0.1 \, \text{cal/g}^\circ \text{C} \)

2. **Set Up the Heat Transfer Equation:**
The heat lost by the water will be equal to the heat gained by the brass vessel:
\[ m_w \cdot c_w \cdot (T_w - T_f) = m_v \cdot c_v \cdot (T_f - T_v) \]
where \( T_f \) is the final temperature.

3. **Substitute the Known Values:**
\[ 20 \cdot 1 \cdot (100 - T_f) = 100 \cdot 0.1 \cdot (T_f - 0) \]
Simplifying this gives:
\[ 20(100 - T_f) = 10T_f \]

4. **Expand and Rearrange the Equation:**
\[ 2000 - 20T_f = 10T_f \]
Combine like terms:
\[ 2000 = 20T_f + 10T_f \]
\[ 2000 = 30T_f \]

5. **Solve for \( T_f \):**
\[ T_f = \frac{2000}{30} = \frac{200}{3} \approx 66.67^\circ \text{C} \]

### Final Answer:
The resulting temperature when 20 g of boiling water is poured into the ice-cold brass vessel is approximately \( 66.67^\circ \text{C} \).