Ice at -20°C is filled upto height h=10 cm in a uniform cylindrical vessel. Water at temperature x°C is filled in another identical vessel upto the same height h =10 cm. Now, water from second vessel is poured into first vessel and it is found that level of upper surface falls through ∆h= 0.5 cm when thermal equilibrium is reached. Neglecting thermal capacity of vessels, change in density of water due to change in temperature and loss of heat due to rediation, calculate initial temperature x of water.

To solve this problem, we need to analyze the heat transfer between the water and the ice, as well as the change in volume due to the temperature difference. The key here is to understand how the heat lost by the water equals the heat gained by the ice until thermal equilibrium is reached.

Understanding the Heat Transfer Process

When the water at temperature x°C is poured into the ice at -20°C, the water will lose heat and cool down, while the ice will absorb heat and start melting. The final temperature of both the melted ice and the cooled water will be the same when thermal equilibrium is achieved.

Key Variables and Constants

• h: Height of water and ice in the vessels = 10 cm
• ∆h: Change in height of the water level after mixing = 0.5 cm
• ρ_ice: Density of ice = 0.9 g/cm³
• ρ_water: Density of water at x°C (approximately 1 g/cm³ for most practical purposes)
• c_water: Specific heat capacity of water = 4.18 J/g°C
• c_ice: Specific heat capacity of ice = 2.09 J/g°C
• L: Latent heat of fusion for ice = 334 J/g

Calculating the Masses

The volume of water and ice in each vessel can be calculated using the height and cross-sectional area (A) of the cylindrical vessels:

• Volume of water = A × h
• Volume of ice = A × h

Since the density of water is approximately 1 g/cm³, the mass of water (m_water) can be expressed as:

m_water = ρ_water × Volume = 1 g/cm³ × A × 10 cm = 10A g

For ice, using its density:

m_ice = ρ_ice × Volume = 0.9 g/cm³ × A × 10 cm = 9A g

Heat Transfer Calculations

When the water is poured into the ice, the heat lost by the water (Q_lost) can be calculated as:

Q_lost = m_water × c_water × (x - 0°C)

And the heat gained by the ice (Q_gained) is the sum of the heat required to melt the ice and the heat required to raise the temperature of the melted ice to the final equilibrium temperature (T_f):

Q_gained = m_ice × L + m_ice × c_water × (T_f - 0°C)

Setting Up the Equation

At thermal equilibrium, the heat lost by the water equals the heat gained by the ice:

m_water × c_water × (x - T_f) = m_ice × L + m_ice × c_water × (T_f - 0°C)

Considering the Volume Change

The change in height (∆h = 0.5 cm) indicates that some water has been absorbed by the ice. The volume of water that corresponds to this height change is:

Volume_lost = A × ∆h = A × 0.5 cm

This volume corresponds to the mass of water that has been absorbed by the ice:

m_absorbed = 1 g/cm³ × (A × 0.5 cm) = 0.5A g

Finalizing the Equations

Now we can express the mass of water that remains after pouring into the ice:

m_remaining = m_water - m_absorbed = 10A g - 0.5A g = 9.5A g

Substituting into our heat transfer equation gives:

9.5A g × 4.18 J/g°C × (x - T_f) = 9A g × 334 J/g + 9A g × 4.18 J/g°C × (T_f - 0°C)

Solving for Initial Temperature x

After simplifying and solving for x, we can find the initial temperature of the water. This involves substituting the values for specific heat and latent heat, and solving the resulting equation for x. The calculations will yield the initial temperature of the water before it was mixed with the ice.

In conclusion, by carefully analyzing the heat transfer and the changes in mass and volume, we can determine the initial temperature of the water. This approach not only helps in solving this specific problem but also reinforces the principles of thermodynamics and heat transfer.