What mass (in grams) of steam at 100°C must be mixed with 340 g of ice at its melting point, in a thermally insulated container, to produce liquid water at 12.0°C? The specific heat of water is 4186 J/kg·K. The latent heat of fusion is 333 kJ/kg, and the latent heat of vaporization is 2256 kJ/kg.
To solve this problem, we need to go step by step and calculate the energy required to change the state of each component (ice and steam) and then calculate the final temperature.
Step 1: Calculate the energy required to melt the ice.
The latent heat of fusion (or melting) is the energy required to convert ice at its melting point to liquid water at the same temperature. The given value is 333 kJ/kg.
The mass of ice is 340 g, so we need to find the total energy required to melt this ice:
Energy required = mass of ice * latent heat of fusion
Energy required = 340 g * (333 kJ/kg)
Energy required = 340g * (333,000 J/kg)
Energy required = 113,220,000 J
Step 2: Calculate the energy required to raise the temperature of the water from the melting point to 12.0°C.
To do this, we need to use the specific heat capacity of water, which is 4186 J/kg·K.
The mass of water after melting the ice will be the same as the mass of the ice (340 g) since there is no change in the amount of water.
Energy required = mass of water * specific heat capacity * change in temperature
Energy required = 340 g * 4186 J/kg·K * (12.0°C - 0°C)
Energy required = 340 g * 4186 J/kg·K * 12.0°C
Energy required = 1,685,952.8 J
Step 3: Calculate the energy released by the condensation of the steam.
The latent heat of vaporization is the energy released when steam condenses to water. The given value is 2256 kJ/kg.
We need to find the mass of steam required to release the same amount of energy as calculated in step 2.
Energy released = mass of steam * latent heat of vaporization
Energy released = mass of steam * (2256 kJ/kg)
Energy released = mass of steam * (2256,000 J/kg)
Since the energy required in step 2 and the energy released in this step are the same, we can equate them:
1,685,952.8 J = mass of steam * (2256,000 J/kg)
Solving for the mass of steam:
mass of steam = 1,685,952.8 J / (2256,000 J/kg)
mass of steam = 0.748 kg
Step 4: Convert the mass of steam to grams:
mass of steam = 0.748 kg * 1000 g/kg
mass of steam = 748 g
Therefore, you would need approximately 748 grams of steam at 100°C to mix with 340 g of ice to produce liquid water at 12.0°C.