Kilocalories to condense 82g of steam at 100 degress C, cool the liquid and freeze it at 0 degress C.
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To calculate the number of kilocalories required, we need to break down the process into several steps:
Step 1: Condensing steam at 100 degrees Celsius to liquid water at 100 degrees Celsius.
Step 2: Cooling the liquid water from 100 degrees Celsius to 0 degrees Celsius.
Step 3: Freezing the liquid water at 0 degrees Celsius into ice.
For each step, we need to consider the specific heat capacity and latent heat of the substance involved. Additionally, we need to account for any phase changes that occur.
1. Condensing steam to liquid water:
The specific heat capacity of water is 1 cal/g°C, which can be converted to 1 kcal/kg°C for convenience. Since we have 82 grams of steam, the calculation can be done as follows:
Q1 = mass (m) * specific heat capacity (c) * ΔT
Q1 = 82g * 1 kcal/kg°C * (100°C - 100°C)
Q1 = 0 kcal
As there is no change in temperature during this phase change, no kilocalories are required.
2. Cooling water from 100°C to 0°C:
The specific heat capacity of water is 1 kcal/kg°C. Again, considering 82 grams of liquid water, the calculation is:
Q2 = mass (m) * specific heat capacity (c) * ΔT
Q2 = 82g * 1 kcal/kg°C * (0°C - 100°C)
Q2 = -820 kcal
Note that the negative sign is because heat is being removed from the water to cool it down.
3. Freezing water at 0°C into ice:
The latent heat of fusion for water is 80 kcal/g. Therefore, for 82 grams of water:
Q3 = mass (m) * latent heat of fusion (L)
Q3 = 82g * 80 kcal/g
Q3 = 6560 kcal
As freezing water into ice is a phase change, no change in temperature occurs.
Now, to get the total kilocalories required, we sum up the values from each step:
Total kilocalories = Q1 + Q2 + Q3 = 0 kcal - 820 kcal + 6560 kcal
Total kilocalories = 5740 kcal
Therefore, it required approximately 5740 kilocalories to condense 82 grams of steam at 100 degrees Celsius, cool the liquid to 0 degrees Celsius, and freeze it into ice at 0 degrees Celsius.