Calculating Total Water Mass in a Heat Transfer Scenario

This problem involves the principles of heat transfer, specifically focusing on phase change (steam condensing) and temperature change of water. We need to determine the total mass of water present after mixing steam with cooler water and reaching thermal equilibrium.

Understanding the Physics Principles

The core principle is that heat lost by the hotter substance (steam) equals the heat gained by the colder substance (water) until they reach a common final temperature. This is based on the law of conservation of energy.

• Heat Gained ($Q_{gain}$): The initial water absorbs heat to increase its temperature. The formula is $Q = m \times c \times \Delta T$, where $m$ is mass, $c$ is specific heat capacity, and $\Delta T$ is the change in temperature.
• Heat Lost ($Q_{lost}$): The steam loses heat in two stages:
  1. Condensation: Steam at 100°C turns into water at 100°C. The formula is $Q = m \times L_v$, where $L_v$ is the latent heat of vaporization.
  2. Cooling: The condensed water at 100°C cools down to the final equilibrium temperature. The formula is $Q = m \times c \times \Delta T$.

Given Data and Constants

• Initial mass of water ($m_w$): 60 g
• Initial temperature of water ($T_i$): 10°C
• Initial temperature of steam ($T_s$): 100°C
• Final equilibrium temperature ($T_f$): 40°C
• Specific heat capacity of water ($c_w$): Assumed to be 1 cal/g°C (a standard value often used in such problems).
• Latent heat of vaporization of steam ($L_v$): Assumed to be 540 cal/g (a standard value).

Step-by-Step Calculation

Step 1: Calculate Heat Gained by Initial Water

The initial 60 g of water heats up from 10°C to 40°C.

Using the formula $Q_{gain} = m_w \times c_w \times (T_f - T_i)$:
$Q_{gain} = 60 \text{ g} \times 1 \text{ cal/g°C} \times (40°C - 10°C)$
$Q_{gain} = 60 \times 1 \times 30$
$Q_{gain} = 1800$ calories

Step 2: Calculate Heat Lost by Steam

Let $m_s$ be the mass of steam added. The steam condenses and then cools.

Heat lost during condensation ($Q_{cond}$):
$Q_{cond} = m_s \times L_v$
$Q_{cond} = m_s \times 540$ calories

Heat lost during cooling of condensed steam (from 100°C to 40°C):
$Q_{cool} = m_s \times c_w \times (T_s - T_f)$
$Q_{cool} = m_s \times 1 \text{ cal/g°C} \times (100°C - 40°C)$
$Q_{cool} = m_s \times 1 \times 60$
$Q_{cool} = 60 m_s$ calories

Total heat lost by steam ($Q_{lost}$):
$Q_{lost} = Q_{cond} + Q_{cool}$
$Q_{lost} = 540 m_s + 60 m_s$
$Q_{lost} = 600 m_s$ calories

Step 3: Equate Heat Gained and Heat Lost

According to the principle of heat exchange: $Q_{gain} = Q_{lost}$.

$1800 \text{ calories} = 600 m_s \text{ calories}$
Solving for $m_s$:
$m_s = \frac{1800}{600}$
$m_s = 3$ grams

This means 3 grams of steam condensed into water.

Step 4: Calculate the Total Mass of Water

The total mass of water at 40°C is the sum of the initial water mass and the mass of steam that condensed.

Total Mass = Initial water mass + Mass of condensed steam
Total Mass = $m_w + m_s$
Total Mass = 60 g + 3 g
Total Mass = 63 g

Conclusion

After the steam is passed into the water, the final mixture consists of the original water plus the condensed steam, resulting in a total mass of 63 grams at 40°C.