Calculating Heat of Vaporization for Solid CO2 in Water

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Homework Help Overview

The discussion revolves around calculating the heat of vaporization for solid carbon dioxide (dry ice) when it is placed in water. The scenario involves 125 grams of dry ice in 500 grams of water at an initial temperature of 66 degrees Celsius, which cools down to 29 degrees Celsius after the dry ice sublimates.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between heat lost by the water and heat gained by the sublimating dry ice, referencing the principle of conservation of energy (Qlost = Qgain). There are discussions about the specific processes involved in sublimation and the heat required for the temperature change of the water.

Discussion Status

Some participants have provided guidance on considering the system as adiabatic and have prompted others to think critically about the heat transfer involved. There is ongoing exploration of the temperature change and its implications for calculating the heat of vaporization.

Contextual Notes

Participants are navigating the problem with an emphasis on understanding the heat transfer dynamics, and there is a noted confusion regarding the application of specific heat and the heat of sublimation in the context of the problem.

nsw
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Help!---heat Of Vaporization

I could really use some help with this please!

125 grams of dry ice (solid CO2)is dropped into a beaker containg 500 grams of 66 degree celsius water. The dry ice converts directly to gas, leaving the solution. When the dry ice is gone, the final temperature of the water is 29 degrees celsius. What is the heat of vaporization of solid CO2?
 
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Homework is handled on a "you show us what you've done, and we'll point you in the direction you need to go." Take a couple wild swings --- think about the initial state (dry ice + water) and the final state, and what has happened to get from initial to final.
 
heat of vaporization

Doesn't it have to do with the Qlost=Qgain
 
I said, "...wild swing..." For a first guess, you're not doing too badly --- at the same time, very few things don't have anything to do with Ql and Qg.

You are evaporating carbon dioxide; the specific process is transition from solid phase to vapor ("sublimation" for the anal-retentive). This process requires heat, specifically, the "heat of vaporization" you used to title the thread. What do you have available as a heat source for this process in this problem?
 
Consider the dry ice and the water to comprise an adiabatic (closed with respect to heat) system, so that the loss of heat (defined in terms of calories) from the water equals the heat needed to sublimate the CO2 mass.
 
Sorry, but i am still confused about this problem.
 
ok what is Qlost & Qgain Write them in terms of mass. specific etc, Heat of sublimation etc
 
Q=mass*specific heat*change in temp.

But how does this help me?
 
Reread the problem: does it state anything about a temperature change?
 
  • #10
yes it does talk about temperature change
 
  • #11
What temperature change?
 
  • #12
The temperature change is of the fluid in the beaker which goes from 66 to 29 degrees celsius.
 
  • #13
How much heat is required for this temperature change?
 

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