Evaporation and Condensation

[92EG6]

Evaporation and Condensation...

If 58 grams of hot water at 71°C is poured into a cavity in a very large block of ice at 0°C, what will be the final temperature of the water in the cavity? in °C


How much ice must melt in order to cool the hot water down to this temperature? in grams

is there an equation for this?

 

[carlz]

There's a formula for this involving latent heat. I kinda forgot the details but there needs to be more information.

How big is the very large block of ice ?? I think the temperature would be 0 degrees celcius.

Anyone ?

 

[quicknote]

I would like to explain the process of evaporation and condensation in this scenario. When the hot water at 71°C is poured into the cavity in the ice block, the heat from the water will cause some of the molecules at the surface of the water to gain enough energy to escape into the air as water vapor. This process is known as evaporation. As more and more molecules escape, the water will start to cool down.

At the same time, the surrounding ice block will absorb the heat from the water, causing some of the ice to melt. This process is known as condensation. As the water vapor molecules come into contact with the cold ice, they lose energy and turn back into liquid water. This process continues until the water reaches equilibrium with the ice and both are at the same temperature.

To determine the final temperature of the water in the cavity, we can use the principle of conservation of energy. We know that the heat energy lost by the hot water will be gained by the ice and the water vapor. Therefore, we can set up an equation:

Heat lost by hot water = Heat gained by ice + Heat gained by water vapor

We can express this in terms of temperature and mass using the specific heat capacity (c) and the change in temperature (ΔT):

m_hotwater x c_hotwater x ΔT_hotwater = m_ice x c_ice x ΔT_ice + m_vapor x c_vapor x ΔT_vapor

We know the initial temperature of the hot water (71°C) and the final temperature of the water and ice (0°C). We also know the specific heat capacities of water and ice, and we can assume that the mass of the water vapor is negligible compared to the water and ice. Therefore, we can solve for the final temperature of the water in the cavity:

m_hotwater x c_hotwater x (71°C - T_final) = m_ice x c_ice x (T_final - 0°C)

Simplifying the equation, we get:

71m_hotwater x c_hotwater = T_final(m_hotwater x c_hotwater + m_ice x c_ice)

Solving for T_final, we get:

T_final = 71m_hotwater x c_hotwater / (m_hotwater x c_hotwater + m_ice x c_ice)

Plugging in the values, we get:

T_final = 71