If 5.00 g of steam (H2O)(g) at 100 C condenses to liquid water (H2O)(l) that is cooled to 50.0 C, how much heat is leberated in calories and kilocalories?
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To calculate the amount of heat liberated when steam condenses to liquid water and then cools down, we need to consider two processes: the condensation of steam to liquid water and the subsequent cooling of the liquid water.
First, let's calculate the heat released during the condensation process:
1. Determine the heat required to condense 1 gram of steam at 100°C:
- The enthalpy of vaporization (ΔHvap) of water is approximately 540 calories/gram at 100°C.
2. Calculate the heat released when 5.00 g of steam condenses to liquid water:
- Multiply the mass of the steam (5.00 g) by the enthalpy of vaporization (540 cal/g) to find the heat released during condensation in calories.
Next, let's calculate the heat released during the subsequent cooling process:
3. Determine the specific heat capacity (C) of water:
- The specific heat capacity of water is approximately 1 calorie/gram°C.
4. Calculate the heat released when the liquid water cools from 100°C to 50.0°C:
- Multiply the mass of the liquid water (which is now 5.00 g since it is the same substance) by the temperature change (50.0°C) and the specific heat capacity of water (1 cal/g°C) to find the heat released during cooling in calories.
Finally, let's convert the heat released from calories to kilocalories:
5. Convert calories to kilocalories:
- Divide the heat released in calories by 1000 to convert to kilocalories.
By following these steps, you can determine the amount of heat liberated both in calories and kilocalories for the given scenario.