Dry Ice. At standard atmospheric pressure, the solid form of carbon dioxide called “dry ice” undergoes a phase change not to a liquid but straight to a gas. This process is called sublimation, and like other phase transitions heat energy is required. In this case it is the latent heat of sublimation, which for carbon dioxide is 573 kJ/kg at the sublimation temperature of ­78.5 °C. If 118 g of dry ice is dropped into 0.51 liters water at room temperature (20 °C), how much of the water will turn to ice by the time all the dry ice has sublimated? Note that when the dry ice becomes a gas, the gas bubbles out of the system.

To determine how much of the water will turn to ice when the dry ice has sublimated, we need to consider the heat transfer between the dry ice and the water.

First, let's find the amount of heat energy released during the sublimation of the dry ice. We have 118 g of dry ice. The latent heat of sublimation for carbon dioxide is 573 kJ/kg.

To find the total heat energy released (Q), we can use the formula:

Q = m × L

Where:
Q = heat energy released during sublimation (in J)
m = mass of dry ice (in kg)
L = latent heat of sublimation (573000 J/g for carbon dioxide)

Converting the mass of dry ice to kg:

m = 118 g / 1000 = 0.118 kg

Now, let's calculate the heat energy released:

Q = 0.118 kg × 573000 J/kg = 67734 J

Next, we need to find the amount of heat energy required to convert the water at room temperature (20 °C) to ice at 0 °C. This can be calculated using the formula:

Q = m × C × ΔT

Where:
Q = heat energy required (in J)
m = mass of water (in kg)
C = specific heat capacity of water (4190 J/kg°C)
ΔT = change in temperature (in °C)

The mass of water is given as 0.51 liters. To convert to kg:

1 liter of water = 1 kg

So, the mass of water is:

m = 0.51 kg

The change in temperature is:

ΔT = 0 °C - 20 °C = -20 °C

Now, let's calculate the heat energy required:

Q = 0.51 kg × 4190 J/kg°C × -20°C = -42492 J

Note that the negative sign is used because heat is being removed from the water during the phase change.

Now, let's determine how much of the water will turn to ice. The heat energy released by the dry ice must be equal to the heat energy required to convert the water to ice:

Q released by dry ice = Q required for water

67734 J = -42492 J

The negative sign indicates heat transfer from the water to the dry ice. However, since we are interested in the amount of water turned to ice, we can ignore the negative sign.

To find the mass of water turned into ice, we can rearrange the formula:

m = Q / ΔH

Where:
m = mass of water turned into ice (in kg)
Q = heat energy released by the dry ice (in J)
ΔH = enthalpy of fusion for water (334000 J/kg)

Now, let's calculate the mass of water turned into ice:

m = 67734 J / 334000 J/kg = 0.203 kg

Therefore, approximately 203 grams of water will turn to ice when all the dry ice has sublimated.