-Steam at 100 ¡ãC is mixed with 166.4 g of ice at ¨C32.8 ¡ãC, in a thermally insulated container, to produce water at 44.6 ¡ãC. Ignore any heat absorption by the container.

-Cwater = 4186. J/(kg ¡ãC)
-Cice = 2090. J/(kg ¡ãC)
-Lf,water = 3.33 ¡Á 105 J/kg
-Lv,water = 2.26 ¡Á 106 J/kg
Also,energy is required to bring all the ice up to 0 ¡ãC=11400J; energy is required to melt the ice into water at 0 ¡ãC=55400J; energy required to raise the temperature of this melted water to 44.6 ¡ãC =31100J;
energy supplied by the steam to change the state of 166.4 g of ice at ¨C32.8 ¡ãC to water at 44.6 ¡ãC =97884J

Q1) What is the final mass of water in the cup at 44.6 ¡ãC??

To solve this question, we need to apply the principles of heat transfer and conservation of energy.

Step 1: Calculate the heat absorbed by the ice to reach 0°C.
The amount of heat absorbed by the ice can be calculated using the formula Q = m * C * ΔT, where Q is the heat absorbed, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

The mass of ice is given as 166.4 g, the specific heat capacity of ice is 2090 J/(kg°C), and the change in temperature is from -32.8°C to 0°C.

Converting the mass to kilograms:
m_ice = 166.4 g = 0.1664 kg

Calculating the heat absorbed:
Q_ice = m_ice * C_ice * ΔT_ice
= 0.1664 kg * 2090 J/(kg°C) * (0 - (-32.8)°C)
= 0.1664 kg * 2090 J/(kg°C) * 32.8°C

Step 2: Calculate the heat absorbed to melt the ice into water at 0°C.
The heat required to melt the ice can be calculated using the formula Q = m * Lf, where Q is the heat absorbed, m is the mass, and Lf is the latent heat of fusion.

The mass of ice remains unchanged at 166.4 g.

Calculating the heat absorbed:
Q_melt = m_ice * Lf
= 0.1664 kg * 3.33 × 10^5 J/kg

Step 3: Calculate the heat absorbed to raise the temperature of the water from 0°C to 44.6°C.
The amount of heat absorbed by the water can be calculated using the formula Q = m * C * ΔT, where Q is the heat absorbed, m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.

As the final temperature is 44.6°C, we do not need to consider any phase changes.

The mass of water can be calculated by subtracting the mass of ice from the total mass of the mixture.
m_water = Total mass - mass of ice

Total mass = mass of steam = mass of steam / density of steam
The density of steam at 100°C is approximately 0.6 kg/m³.
Using the formula: Mass = Volume * Density
Mass_steam = Volume_steam * Density_steam
= (mass_steam / density_steam) * density_steam
= mass_steam

Therefore, Mass_water = mass_steam - mass_ice

Now, calculating the heat absorbed:
Q_water = m_water * C_water * ΔT_water
= (mass_steam - mass_ice) * C_water * (44.6 - 0)°C

Step 4: Calculate the total heat absorbed by the system.
The total heat absorbed is equal to the energy supplied by the steam to change the state of the ice to water, combining all the calculated values.

Total heat absorbed = Q_ice + Q_melt + Q_water
= 0.1664 kg * 2090 J/(kg°C) * 32.8°C + 0.1664 kg * 3.33 × 10^5 J/kg + (mass_steam - mass_ice) * C_water * (44.6 - 0)°C

Now, equate this total heat absorbed to the energy supplied by the steam to change the state of the ice to water:

Total heat absorbed = Energy supplied by steam
Q_ice + Q_melt + Q_water = 97884 J

Step 5: Solve for the unknown, which is the final mass of water.
Rearrange the equation to isolate the final mass of water:

mass_water = (Energy supplied by steam - (Q_ice + Q_melt)) / (C_water * ΔT_water) + mass_ice

Substitute the known values into the equation and calculate:

mass_water = (97884 J - (0.1664 kg * 2090 J/(kg°C) * 32.8°C + 0.1664 kg * 3.33 × 10^5 J/kg)) / (4186 J/(kg°C) * 44.6°C) + 0.1664 kg

Simplifying the equation will give you the final mass of water in the cup at 44.6°C.