What mass (in grams) of steam at 100°C must be mixed with 341 g of ice at its melting point, in a thermally insulated container, to produce liquid water at 57.0°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.
add the heats gained, and set the sum to zero.
Heat melting ice + heat raising temp ice to 57C + heat gained condensing steam + heat lowering condensate temp to 57=0
.341*333J+.341*4146(57-0)+M*2256+M(4146)(57-100)=0
solve for Mass (in kilograms). Change it to grams for your answer.
To solve this problem, we need to calculate the amount of heat transferred during the phase changes and the temperature changes. Here's how you can approach it:
1. Calculate the heat required to melt the ice at its melting point (0°C) to liquid water at 0°C:
Q1 = mass of ice x latent heat of fusion
Given:
- Mass of ice = 341 g
- Latent heat of fusion = 333 kJ/kg
- 1 kg = 1000 g
Convert the mass of ice to kg:
mass of ice = 341 g ÷ 1000 g/kg
Substitute the values into the equation:
Q1 = (mass of ice ÷ 1000 kg/g) x (333,000 J/kg)
2. Calculate the heat required to heat the melted ice from 0°C to 57.0°C:
Q2 = mass of liquid water x specific heat x temperature change
Given:
- Specific heat of water = 4186 J/kg·K
- Temperature change = 57.0°C - 0°C
Substitute the values into the equation:
Q2 = (mass of liquid water) x (4186 J/kg·K) x (57.0°C - 0°C)
3. Calculate the heat required to convert the liquid water at 57.0°C to steam at 100°C:
Q3 = mass of steam x latent heat of vaporization
Given:
- Latent heat of vaporization = 2256 kJ/kg
Convert the latent heat of vaporization to J/kg:
Latent heat of vaporization = 2256 kJ/kg x 1000 J/kJ
Substitute the values into the equation:
Q3 = (mass of steam) x (2,256,000 J/kg)
4. The total heat gained by the system is equal to the sum of the three heats:
Q total = Q1 + Q2 + Q3
5. Since the container is insulated, the total heat gained by the system equals the total heat lost by the system. The heat lost by the steam at 100°C corresponds to its temperature drop to 57.0°C:
Q lost = mass of steam x specific heat x temperature change
Using the same specific heat as before and the temperature change of 43.0°C:
Q lost = (mass of steam) x (4186 J/kg·K) x (100°C - 57.0°C)
6. Set the total heat gained equal to the total heat lost and solve for the mass of steam:
Q total = Q lost
Once you solve for the mass of steam, you will have found the answer to the question.