The Question is from neet-physics thermal-properties-of-matter calorimetry, and it was created by the author P.NARASIMHA

Question

Steam is passed into $22g$ of water at ${20^{\rm{o}}}{\rm{C}}$. The mass of water that will be present when the water acquires a temperature of ${90^{\rm{o}}}{\rm{C}}$ (Latent heat of steam is ) is $540\,\,cal/g)$

P.NARASIMHA · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [A]
## CONCEPT FRAMEWORK
**Primary Concept**: Heat transfer and phase change.
**Related Topics**: Latent heat of vaporization, specific heat capacity, calorimetry.
**Prerequisites**: Understanding of heat transfer mechanisms, specific heat, and latent heat.
## SOLUTION PATHWAY
**Given Information**:
   - Mass of water (m_w) = 22 g
   - Initial temperature of water (T_wi) = 20°C
   - Final temperature of water (T_wf) = 90°C
   - Latent heat of steam (L_v) = 540 cal/g
   - Specific heat of water (c_w) = 1 cal/g°C
**Method & Steps**:
  1. Calculate the heat gained by the water:
     Reasoning: The water absorbs heat to raise its temperature.
     Formula/Rule: Q_water = m_w * c_w * (T_wf - T_wi)
  2. Calculate the heat lost by the steam:
     Reasoning: Steam condenses into water and then cools down to 90°C. Let m_s be the mass of steam.
     Formula/Rule: Q_steam = m_s * L_v + m_s * c_w * (100 - 90)
  3. Equate heat gained by water and heat lost by steam, and solve for m_s:
     Reasoning: Heat gained by water is equal to heat lost by steam
     Formula/Rule: Q_water = Q_steam
  4. Add m_s to the initial mass of water to obtain total mass.
**Working**:
  1.  Q_water = 22 g * 1 cal/g°C * (90°C - 20°C) = 22 * 70 = 1540 cal
  2.  Q_steam = m_s * 540 cal/g + m_s * 1 cal/g°C * (100°C - 90°C) = 540 m_s + 10 m_s = 550 m_s cal
  3.  1540 = 550 m_s 
       m_s = 1540/550 = 2.8 g
  4.  Total mass = 22 g + 2.8 g = 24.8 g
**Result**: Total mass of water = 24.8 g
**Verification**: The answer seems reasonable as the steam mass needed is less compared to the original water mass.
## OPTION ANALYSIS
[A] ✓: Correct answer calculated using the heat transfer and phase change concepts.
[B] :x: Incorrect calculation.
[C] :x: Incorrect calculation.
[D] :x: Incorrect calculation.
## LEARNING AIDS
**Common Mistakes**:
  1. [Error 1]: Forgetting to include the heat released by the condensation of steam.
   How to avoid: Always consider all the heat transfer processes and phase change.
  2. [Error 2]: Using the wrong temperature difference for the water.
   How to avoid: Carefully consider temperature changes in each step.
**Quick Tips**:
  1. [Helpful shortcut]: Always remember to include all heat transfer processes in a calorimetry problem.
  2. [Memory aid]: Q=mcΔT for temperature change and Q=mL for phase change.

P.NARASIMHA · Answer

The correct Option is "$24.8g$"

P.NARASIMHA · Answer

$24g$

P.NARASIMHA · Answer

$36.6g$

P.NARASIMHA · Answer

$30g$

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