In a calorimeter of water equivalent 20 g, water of mass 1.1 kg is taken at 288 K temperature. If steam at temperature 373 K is passed through it and temperature of water increases by 6.5°C then the mass of steam condensed is

To determine the mass of steam that condensed in the calorimeter, we can use the principle of conservation of energy, which states that the heat lost by the steam will equal the heat gained by the water and the calorimeter. Let’s break down the problem step by step.

Understanding the System

We have a calorimeter with a water equivalent of 20 g and 1.1 kg of water at an initial temperature of 288 K. When steam at 373 K is introduced, the temperature of the water increases by 6.5°C. We want to calculate the mass of steam that condensed during this process.

Key Information

• Mass of water (m_water) = 1.1 kg = 1100 g
• Initial temperature of water (T_initial) = 288 K
• Final temperature of water (T_final) = 288 K + 6.5°C = 294.5 K
• Water equivalent of the calorimeter (W_calorimeter) = 20 g
• Latent heat of vaporization of steam (L) = 540 cal/g
• Specific heat of water (c_water) = 1 cal/g°C

Calculating Heat Gained by Water and Calorimeter

The total heat gained by the water and the calorimeter can be calculated using the formula:

Q_gained = (m_water + W_calorimeter) × c_water × ΔT

Where:

• ΔT = T_final - T_initial = 294.5 K - 288 K = 6.5°C

Substituting in the values:

Q_gained = (1100 g + 20 g) × 1 cal/g°C × 6.5°C

Q_gained = 1120 g × 6.5 cal/g = 7280 cal

Calculating Heat Lost by Steam

The heat lost by the steam as it condenses and cools can be divided into two parts:

• The heat released during condensation
• The heat released as the resulting water cools to the final temperature

Let the mass of steam condensed be m_steam. The heat lost during condensation can be calculated as:

Q_condensation = m_steam × L

After condensation, the steam turns into water at 373 K. This water then cools down to the final temperature of 294.5 K, and the heat lost during this cooling can be calculated as:

Q_cooling = m_steam × c_water × (373 K - 294.5 K)

Where the temperature difference is (373 K - 294.5 K) = 78.5°C.

Thus, the total heat lost by the steam can be expressed as:

Q_lost = Q_condensation + Q_cooling

Q_lost = m_steam × L + m_steam × c_water × (373 K - 294.5 K)

Setting Up the Equation

Since the heat gained by the water and calorimeter equals the heat lost by the steam, we can set up the following equation:

Q_gained = Q_lost

7280 cal = m_steam × 540 cal/g + m_steam × 1 cal/g × 78.5°C

Combining the terms gives:

7280 cal = m_steam × (540 + 78.5) cal/g

7280 cal = m_steam × 618.5 cal/g

Solving for Mass of Steam

Now, we can solve for m_steam:

m_steam = 7280 cal / 618.5 cal/g

m_steam ≈ 11.77 g

Final Result

The mass of steam that condensed in the calorimeter is approximately 11.77 grams. This calculation demonstrates how energy transfer occurs in thermal systems and highlights the applications of the laws of thermodynamics in practical scenarios.