Final answer:
To find the total heat released when 175 g of steam condenses at 100 °C and the water cools to 25 °C, we calculate the heat from the condensation using the latent heat of vaporization and from cooling using the specific heat capacity. The total heat released is then the sum of the two calculated values, giving 449.9625 kJ.
Step-by-step explanation:
To calculate the joules released when 175 g of steam condenses at 100 °C and then the liquid cools to 25.0 °C, we need to consider two separate processes: the condensation of steam at 100 °C and the cooling of the resulting liquid water from 100 °C to 25.0 °C.
For the first process, the heat released during the condensation of steam can be computed using the latent heat of vaporization of water, which is 2257 kJ/kg at 100 °C. The mass of steam is given in grams and must be converted to kilograms to use this constant:
- 175 g = 0.175 kg
- Heat released during condensation (Q_condensation) = mass × latent heat of vaporization
- Q_condensation = 0.175 kg × 2257 kJ/kg
- Q_condensation = 395.475 kJ
For the second process, the heat released during the cooling of liquid water is calculated using the specific heat capacity of water, which is approximately 4.18 J/g°C. Since the mass is given in grams, there is no need to convert to kilograms:
- Temperature change (delta T) = 100 °C - 25 °C = 75 °C
- Heat released during cooling (Q_cooling) = mass × specific heat capacity × delta T
- Q_cooling = 175 g × 4.18 J/g°C × 75 °C
- Q_cooling = 54,487.5 J or 54.4875 kJ
Adding the heat from both processes gives the total heat released:
- Total heat released (Q_total) = Q_condensation + Q_cooling
- Q_total = 395.475 kJ + 54.4875 kJ
- Q_total = 449.9625 kJ
Therefore, the total amount of heat released is 449.9625 kJ.