Q. 1kg of ice at 0^o C is mixed with 1 kg of steam at 100^o C . What will be the composition of the system when thermal equilibrium is reached? Given: Latent heat of fusion of ice = 3.36 x 10^5 J/kg and Latent heat of vaporization of water = 2.26 x 10^6 J/kg.I want the method of approach to this problem. I think we must use Q = mL. But we need the composition i.e x gm of steam and y gm of water.So how to solve it.

To tackle the problem of mixing 1 kg of ice at 0°C with 1 kg of steam at 100°C, we need to consider the energy exchanges that occur as the system reaches thermal equilibrium. You're correct that we will use the formula \( Q = mL \), where \( Q \) is the heat energy, \( m \) is the mass, and \( L \) is the latent heat. Let's break down the steps to find the final composition of the system.

Understanding the Energy Transfers

When the ice and steam are mixed, several processes occur:

• The ice will absorb heat and melt into water at 0°C.
• The steam will release heat and condense into water at 100°C.
• Once both substances are in the liquid state, they will reach thermal equilibrium at a common temperature.

Step 1: Calculate the Heat Required to Melt the Ice

The first step is to determine how much heat is needed to convert the ice into water. Using the latent heat of fusion:

Q_ice = m_ice × L_fusion

Here, \( m_{ice} = 1 \, \text{kg} \) and \( L_{fusion} = 3.36 \times 10^5 \, \text{J/kg} \).

So,

Q_ice = 1 \, \text{kg} × 3.36 \times 10^5 \, \text{J/kg} = 3.36 \times 10^5 \, \text{J}

Step 2: Calculate the Heat Released by the Steam

Next, we calculate how much heat is released when the steam condenses into water. Using the latent heat of vaporization:

Q_steam = m_steam × L_vaporization

Here, \( m_{steam} = 1 \, \text{kg} \) and \( L_{vaporization} = 2.26 \times 10^6 \, \text{J/kg} \).

So,

Q_steam = 1 \, \text{kg} × 2.26 \times 10^6 \, \text{J/kg} = 2.26 \times 10^6 \, \text{J}

Step 3: Determine the Final Composition

Now, we need to see if the heat released by the steam is sufficient to melt the ice and possibly raise the temperature of the resulting water. The total heat released by the steam is:

Q_total = Q_steam = 2.26 \times 10^6 \, \text{J}

Comparing the heat required to melt the ice:

Q_ice = 3.36 \times 10^5 \, \text{J}

Since \( Q_{total} > Q_{ice} \), the steam can fully melt the ice. After melting, we have:

• 1 kg of water from melted ice.
• 1 kg of water from condensed steam.

This gives us a total of 2 kg of water at 100°C initially. However, we need to check if all steam condenses or if some remains as vapor.

Step 4: Calculate the Remaining Heat

After melting the ice, the heat used is:

Remaining heat = Q_steam - Q_ice = 2.26 \times 10^6 \, \text{J} - 3.36 \times 10^5 \, \text{J} = 1.92 \times 10^6 \, \text{J}

This remaining heat will be used to raise the temperature of the 2 kg of water formed. The specific heat capacity of water is approximately \( 4.18 \times 10^3 \, \text{J/(kg°C)} \).

Using the formula \( Q = mc\Delta T \), we can find the temperature increase:

1.92 \times 10^6 \, \text{J} = 2 \, \text{kg} × 4.18 \times 10^3 \, \text{J/(kg°C)} × \Delta T

Solving for \( \Delta T \):

ΔT = \frac{1.92 \times 10^6 \, \text{J}}{2 \, \text{kg} × 4.18 \times 10^3 \, \text{J/(kg°C)}} ≈ 229.1°C

Since this temperature increase is not possible (as water cannot exceed 100°C at atmospheric pressure), all steam condenses, and the final equilibrium temperature will be 100°C.

Final Composition

At thermal equilibrium, we have:

• 0 kg of ice (all melted)
• 1 kg of steam (all condensed)
• 2 kg of water (from both melted ice and condensed steam)

Thus, the final composition of the system will be 2 kg of water at 100°C, with no remaining ice or steam. This thorough approach illustrates how energy conservation principles govern the phase changes and temperature equilibria in thermodynamic systems.