What is the final temperature of water

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SUMMARY

The final temperature of water can be calculated using the formula Q = m*C*T, where Q is the heat energy absorbed, m is the mass, C is the specific heat, and T is the change in temperature. For 335 g of water absorbing 2.66 kJ of energy, the specific heat of water is 4.184 J/g°C, leading to a final temperature calculation of approximately 27.1°C. In the second problem, the molar heat of vaporization of bromine (Br2) can be determined using Q = m*L, where 7.28 kJ of heat is released when 40.9 g of Br2 condenses. The boiling point of bromine is around 331 K, which is relevant for understanding the phase change involved.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically heat transfer.
  • Familiarity with the specific heat capacity of water (4.184 J/g°C).
  • Knowledge of latent heat concepts, particularly for phase changes.
  • Basic algebra skills for solving equations involving heat energy.
NEXT STEPS
  • Learn how to apply the heat transfer formula Q = m*C*T in various scenarios.
  • Research the latent heat of vaporization for different substances, including bromine.
  • Explore the concept of phase changes and the significance of boiling points in thermodynamics.
  • Study the specific heat capacities of various liquids and gases for comparative analysis.
USEFUL FOR

Students preparing for chemistry exams, educators teaching thermodynamics, and professionals in fields requiring heat transfer calculations.

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I got achem test tomororw and 2 questions didn't have asnwers for them so i want to make sure of them.

1) If 335 g of water at 24.5 degrees celsius absorbed 2.66 kJ of energy, what is the final temperature of water

*2.66 = 335 * 4.184 * t + 24.5 (or 24.5 - t)? don't know

and


2)7.28 kJ of heat is released when 40.9 g of Br2 (bormine) vapour at 332 K condense to a liquid. What is the molar heat of vaporization of Br2.


i just want to make sure i got the right asnwers, cause these 2 don't have asnwers in our test revision sheet. thanks
 
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For these two problems you need to remember two of the heat energy formulas,
Q = m*C*T
And
Q = m*L
Where Q is the heat energy, m is the mass, C is the specific heat, T is the change in temperature, and L is the Latent heat [of vaporization].

For your first problem, you need to remember to keep the units on your energy straight, or the prefixes atleast.
2.66 kJ of energy are absorbed, but the specific heat your using for water is 4.184 J/g*degree C
So you need to convert one or the other to get the right answer.
So then just solve for the change in temperature.
2660 J = 335 g * 4.184 J/gC * T

For your second problem,
You know the energy given off and the mass of the bromine giving it off, so just solve for the latent heat.
7280 J = 40.9 g * L
 
god bless u mr. jeffey
for the 2nd prob tho, is the 332K insignifcant
dont i have to find out the heat it takes to get to its vap. piont first.
 
Not knowing the boiling point of bromine, I just assumed the temperature given was the boiling point and left it at that.
However, if it is not and the gas must first cool down and then condense, it becomes a multipart problem, and you then also need to know the specific heat of bromine gas.

Upon further inspection,
the boiling point of Br2 is 136 degrees F (331 K or thereabouts), so it is almost insignificant the amount of heat it looses before it condenses.
http://www.jtbaker.com/msds/englishhtml/B3905.htm
 

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