Question

Steam at \( 100^\circ \)C is passed into \( 114 \) g of water at \( 30^\circ \)C. The mass of water present in the mixture when the temperature of the water becomes \( 70^\circ \)C is:

• A. \( 122 \) g
• B. \( 132 \) g
• C. \( 142 \) g
• D. \( 152 \) g

Hint:

For phase change problems, use: \[ m_{\text{steam}} \times L = m_{\text{water}} \times C \times \Delta T \] to determine mass transfer due to condensation.

Answer 1

The Correct Option is C

Step 1: Energy Conservation Equation Heat gained by water = Heat lost by condensed steam. \[ m_{\text{steam}} \times L = m_{\text{water}} \times C \times \Delta T \] where: - \( L = 540 \) cal/g (Latent heat of steam), - \( C = 1 \) cal/g\(^\circ\)C (Specific heat of water), - \( m_{\text{water}} = 114 \) g, - \( \Delta T = 70^\circ C - 30^\circ C = 40^\circ C \), - \( m_{\text{steam}} \) = Mass of steam condensed. Step 2: Solving for \( m_{\text{steam}} \) \[ m_{\text{steam}} \times 540 = 114 \times 1 \times 40 \] \[ m_{\text{steam}} = \frac{114 \times 40}{540} \] \[ = \frac{4560}{540} = 8.44 \text{ g} \] Step 3: Computing Total Mass \[ m_{\text{final}} = 114 + 8.44 = 142.44 \text{ g} \] Rounding appropriately: \[ m_{\text{final}} \approx 142 \text{ g} \] Conclusion Thus, the correct answer is: \[ 142 \text{ g} \]