What mass (in grams) of steam at 100°C must be mixed with 446 g of ice at its melting point, in a thermally insulated container, to produce liquid water at 65.0°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 melt 446 g of ice and heat it to 65 C requires heat transfer to that ice of 446(333 + 65*4186)
Set that number equal to the heat transferred from the steam mass M, which is
M(2256 + 35*4186)
Solve for the single unknown,M, in grams
816Kg
NO
To solve this problem, we can use the principles of energy conservation and the specific heat equation.
First, let's break down the problem and identify the different phases and temperatures involved:
1. We have two substances: steam at 100°C and ice at its melting point (0°C).
2. We want to end up with liquid water at 65.0°C.
Now, let's calculate the energy transfer during each phase change:
1. Melting ice: The latent heat of fusion is the amount of energy required to convert 1 kg of ice at its melting point to 1 kg of water at the same temperature. In our case, the latent heat of fusion is 333 kJ/kg, so the total energy required to melt the ice is (333 kJ/kg) * (446 g / 1000) = 148.518 kJ.
2. Heating the water from 0°C to 65.0°C: To calculate the energy required to heat the water, we'll use the specific heat equation:
Energy = mass * specific heat * temperature change.
Since we have liquid water, the specific heat of water is 4186 J/kg·K. Therefore, the energy required for this phase change is (mass of water in kg) * 4186 J/kg·K * (65.0°C - 0°C).
3. Condensing steam to water: The latent heat of vaporization is the amount of energy released when 1 kg of steam condenses to form water at the same temperature. In our case, the latent heat of vaporization is 2256 kJ/kg. Hence, the energy released during this phase change is (2256 kJ/kg) * (mass of steam in kg).
Next, we'll set up the energy balance equation:
Energy gained by heating the water + Energy released by condensing steam = Energy required to melt the ice
Now, let's calculate the masses and energies involved:
1. Mass of ice = 446 g = 0.446 kg
2. Mass of steam = unknown, let's call it 'm'.
Using the energy balance equation, we can write the equation:
(mass of water in kg) * 4186 J/kg·K * (65.0°C - 0°C) + (mass of steam in kg) * 2256 kJ/kg = 148.518 kJ
Now, we can solve this equation to find the mass of water:
(0.046 kg) * 4186 J/kg·K * (65.0°C - 0°C) + (mass of steam in kg) * 2256 kJ/kg = 148.518 kJ
Simplifying the equation:
0.046 kg * 4186 J/kg·K * 65.0°C + (mass of steam in kg) * 2256 kJ/kg = 148.518 kJ
Solving for the mass of steam:
(mass of steam in kg) * 2256 kJ/kg = 148.518 kJ - (0.046 kg * 4186 J/kg·K * 65.0°C)
(mass of steam in kg) = (148.518 kJ - (0.046 kg * 4186 J/kg·K * 65.0°C)) / 2256 kJ/kg
Now, substitute the known values into the equation and calculate the mass of steam.