"Calculate the amount of heat (kcal) released when 18.0 g (grams) of steam are condensed at 100.0 C and the resulting water is frozen and cooled to 25.0 C."
The only problem that I'm having is the amount of physical and temperature changes that there are. I'm new and still learning all of this. If anyone could please tell me how many physical and temperature changes there are than I can do the rest on my own (the equations). Thank you!
To calculate the amount of heat released in this process, let's break it down into steps:
Step 1: Condensation of Steam
- There is a phase change from steam to liquid water at 100.0 °C.
- The heat released during condensation is given by the equation: Q = m * ΔH_vap, where Q is the heat released, m is the mass of the substance, and ΔH_vap is the heat of vaporization.
- In this case, m = 18.0 g (given) and ΔH_vap = 40.7 kcal/mol (heat of vaporization for water).
- To convert mass to moles, divide by the molar mass of water: 18.0 g / (18.0 g/mol) = 1.0 mol.
- Therefore, Q = 1.0 mol * 40.7 kcal/mol.
Step 2: Freezing of Water
- There is a phase change from liquid water to ice at 0 °C.
- The heat released during freezing is given by the equation: Q = m * ΔH_fus, where ΔH_fus is the heat of fusion.
- The heat of fusion for water is ΔH_fus = 1.44 kcal/mol.
- Since we have 1.0 mol of water from the previous step, Q = 1.0 mol * 1.44 kcal/mol.
Step 3: Cooling of Ice
- The temperature of ice decreases from 0 °C to 25.0 °C.
- The heat released during cooling is given by the equation: Q = m * c * ΔT, where c is the specific heat capacity and ΔT is the change in temperature.
- For ice, the specific heat capacity is c = 0.48 kcal/(mol·°C).
- We need to determine the moles of ice, which is the same as the moles of water in step 1 (1.0 mol).
- ΔT = (25.0 °C - 0 °C) = 25.0 °C.
- Therefore, Q = 1.0 mol * 0.48 kcal/(mol·°C) * 25.0 °C.
To find the total heat released, add up the Q values from all three steps: Q_total = Q1 + Q2 + Q3.
Please note that in this calculation, we assume no heat is lost to the surroundings and that the heat capacities and heat of phase change values are constant.
temp changes:
water from 100 to 0C
ice from Oc to xxxx
Now xxxx. if the water is frozen, it freezes at 0C, so you must mean cooled to -25C.
Physical changes:
steam condensing
water freezing