What mass of steam at 100¡ãC must be mixed with 388 g of ice at its melting point, in a thermally insulated container, to produce liquid water at 45.0¡ãC? 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.
.388 kg ice
melt it and warm it to 45 C
.388 [ 333,000 + 45(4186) ]
= 202,292 Joules into what was ice
= m [2256,000 + 55(4186) ]
= 2,486,230 m
m = 202,292/2,486,230
= .0814 kg
= 81.4 grams
To solve this problem, we need to calculate the amount of heat gained or lost in each step of the process and equate it to the final desired temperature.
Step 1: Heating the ice to 0°C
In this step, we need to calculate the heat gained by the ice to reach its melting point.
Heat gained = mass × specific heat capacity × temperature change
The mass of ice is given as 388 g, the specific heat capacity of water is 4186 J/kg·K, and the temperature change is 0°C - (-273.15°C) = 273.15 K.
So the heat gained in this step is:
Heat gained = 388 g × (1 kg / 1000 g) × 4186 J/kg·K × 273.15 K
Step 2: Melting the ice
In this step, we need to calculate the heat required to melt the ice.
Heat required = mass × latent heat of fusion
The mass is the same as the mass of ice calculated in the previous step. The latent heat of fusion is given as 333 kJ/kg, which needs to be converted to joules.
So the heat required in this step is:
Heat required = 388 g × (1 kg / 1000 g) × (333 kJ/kg × 1000 J/kJ)
Step 3: Heating the water to 45°C
In this step, we need to calculate the heat gained by the water to reach the final temperature.
Heat gained = mass × specific heat capacity × temperature change
The mass of water is what we need to find. The specific heat capacity of water is 4186 J/kg·K, and the temperature change is 45°C - 0°C = 45 K.
So the heat gained in this step is:
Heat gained = mass × 4186 J/kg·K × 45 K
Step 4: Condensing the steam
In this step, we need to calculate the heat released when the steam condenses.
Heat released = mass × latent heat of vaporization
The mass of steam is what we need to find. The latent heat of vaporization is given as 2256 kJ/kg, which needs to be converted to joules.
So the heat released in this step is:
Heat released = mass × (2256 kJ/kg × 1000 J/kJ)
Now, to solve the problem, we know that the heat gained in Step 1 plus the heat required in Step 2 must be equal to the heat gained in Step 3 plus the heat released in Step 4.
(Heat gained in Step 1 + Heat required in Step 2) = (Heat gained in Step 3 + Heat released in Step 4)
388 g × (1 kg / 1000 g) × 4186 J/kg·K × 273.15 K + 388 g × (1 kg / 1000 g) × (333 kJ/kg × 1000 J/kJ) = mass × 4186 J/kg·K × 45 K + mass × (2256 kJ/kg × 1000 J/kJ)
Simplifying and solving for the mass of steam (mass), we get:
mass = [(388 × 1 × 4186 × 273.15) + (388 × 1 × 333)] / [(4186 × 45) + (2256 × 1000)]
mass ≈ 203.67 g
Therefore, approximately 203.67 g of steam at 100°C must be mixed with 388 g of ice at its melting point to produce liquid water at 45.0°C.