given the following info for water specific heat capacity of steam 2.03 j/g degrees calculate the total quanity of heat evolved when 10.0g of steam is condensed,cooled,and frozen to ice at -50 degrees
This is best done in stages. There are just two formulas involved. Remember them.
q1 = heat released on the phase change from vapor (steam) at 100 C to liquid water at 100 C.
q1 = mass steam x heat vaporization. (formula 1)
q2 = heat released on cooling liquid water at 100 C to liquid water at zero C.
q2 = mass water x specific heat water x (Tfinal-Tinitial)[formula 2--not a phase change but moving WITHIN a phase, in this case all liquid].
q3 = heat released on the phase change of liquid water at zero C to solid ice at zero C.
q3 = mass water x heat fusion (formula 1 again except we are freezing it and not condensing it).
q4 = heat released on cooling ice from zero C to -50C. (moving within a phase again)
q4 = mass ice x specific heat ice x (Tfinal-Tinitial) (formula 2 again.)
Total q = sum of individual qs.
A 3.0−L sample of a 6.2M NaCl solution is diluted to 75L .
To calculate the total quantity of heat evolved, we need to break down the process into three steps: condensation, cooling, and freezing.
Step 1: Condensation
To condense steam to water, we need to calculate the heat released using the specific heat capacity of steam. The equation for calculating heat is Q = m * c * ΔT, where Q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
In this case, the mass (m) is 10.0g, specific heat capacity (c) is 2.03 J/g°C (given), and the change in temperature (ΔT) is the difference between the boiling point of water (100°C) and the temperature of the steam. Assuming the steam is at 100°C, ΔT would be 100 - 100 = 0°C.
Therefore, the heat released during condensation is:
Q1 = 10.0g * 2.03 J/g°C * 0°C = 0 J
Step 2: Cooling
Now, we need to calculate the heat released during the cooling process from 100°C to 0°C. We will use the same formula with the specific heat capacity of water instead, as steam has already condensed.
The specific heat capacity of water is usually around 4.18 J/g°C. Assuming the specific heat capacity is the same for this scenario, the heat released during the cooling process is:
Q2 = 10.0g * 4.18 J/g°C * (-100°C) = -4180 J
Note that the negative sign indicates the release of heat.
Step 3: Freezing
Lastly, we need to calculate the heat released during the freezing process from 0°C to -50°C. The specific heat capacity of ice is also around 2.03 J/g°C.
The heat released during the freezing process is:
Q3 = 10.0g * 2.03 J/g°C * (-50°C) = -1015 J
Again, the negative sign indicates the release of heat.
Total Heat Evolved:
To find the total heat evolved, we add up the heats released during each step:
Total Heat Evolved = Q1 + Q2 + Q3 = 0 J + (-4180 J) + (-1015 J) = -5195 J
So, the total quantity of heat evolved when 10.0g of steam is condensed, cooled, and frozen to ice at -50°C is -5195 J (negative sign indicates the release of heat).