How much heat is lost when 12.55 g of zinc vapor at 907 degrees celsius is cooled to 425 degrees celsius?
What's the freezing point and boiling point of Zn? You need those to work the problem.
boiling point- 907 degrees celsius
melting point- 419 degrees celsius
At 407 the Zn metal vapor is at the boiling point. First it must be condensed (at 407C) to become a liquid. Cooled to 425 it hasn't frozen yet so it is still a liquid.
q1 = heat released on condensing the Zn vapor.
q1 = mass Zn vapor x heat vaporization/condensation.
q2 = heat released on cooling to 425 C.
q2 = mass Zn x specific heat Zn LIQUID X (Tfinal-Tinitial)
Total heat released is q1 + q2. I don't know the specific heat of Zn metal liquid.
To determine the amount of heat lost during the cooling process, you can use the specific heat capacity equation:
q = m * c * ∆T
where:
q represents the amount of heat energy transferred,
m is the mass of the substance,
c is the specific heat capacity of the substance, and
∆T is the change in temperature (final temperature - initial temperature).
In this case, you need to find the amount of heat lost as the zinc vapor cools down from 907 degrees Celsius to 425 degrees Celsius.
First, you need to know the specific heat capacity of zinc. The specific heat capacity of zinc is approximately 0.388 J/g°C.
Next, calculate the change in temperature:
∆T = final temperature - initial temperature
∆T = 425°C - 907°C
∆T = -482°C
The negative sign indicates a decrease in temperature.
Now, substitute the values into the formula:
q = m * c * ∆T
q = 12.55 g * 0.388 J/g°C * (-482°C)
Calculate the result:
q = -23,794.68 J
Therefore, approximately 23,794.68 J of heat energy is lost when 12.55 g of zinc vapor at 907 degrees Celsius is cooled to 425 degrees Celsius.