# Chemistry

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"Energy is released when 100 kg of steam at 150 degrees Celsius and standard pressure changes to water at 60 degrees Celsius. Calculate the total energy change".

I used q=mct to find the energy required to change to 100 degrees Celsius. But now what??

• Chemistry - ,

Now you need to add the heat released when steam at 100 changes to liquid water at 100.
q = mass x heat vaporization.
Then mcdelta T to go from 100 to 60.

• Chemistry - ,

Thank you! However, I think I'm making a mistake somewhere:

(100000g)(2.02 J/gC)(100C-150C) +
(100000g)/(18.02g/mol)(40650J/mol) +
(100000g)(4.19J/gC)(60C-100C)
Then divide by 1000 to get to kJ.

I don't seem to be getting the right answer (252*10^3kJ).

• Chemistry - ,

I think the error is one of signs.
If you do it
q1 = mc*delta T and DON'T use Tf-Ti. Use 50.
q2 = your value is ok. Technically, it's a negative sign but the way you have it written it is a positive number and all of the numbers should be added.
q3 = mc*delta T and don't use Tf-Ti. Just just 40.
Then total q = q1 + q2 + q3.

OR, you can simply change the sign of the middle term that you posted to a negative sign and add all the three negative numbers.

• Chemistry - ,

Also, I would use 4.184 J/g*C for the specific heat of water instead of rounding to 4.19

• Chemistry - ,

Thank you so much! Sorry, just one more question. Why is the middle term supposed to be a negative? Is it just a math error...or is it because the water is cooling down (hence negative)?

• Chemistry - ,

Oh, with the data booklet that I'm given, it only shows 4.19.

• Chemistry - ,

ok on the 4.19.
The middle term is negative because it is losing heat. Remember something giving off heat is -. Absorbing heat is +. The mass*C*delta T thing takes care of the sign automatically when you use Tfinal-Tinitial (it makes both of those terms negative BECAUSE Tfinal is first and is the lower temp and Tinitial is higher. Automatic - term when it is losing heat and automatic + term when it is gaining heat. BUT the q = mass x heat vap has no such automatic and we must make q + or - depending upon which way the heat is going. That's why I don't put it together as one long equation because I invariably forget to take care of that middle sign. I calculate them separately and use just delta T. Then I add all of them together as positive numbers (because we are going up the scale and all of the numbers are positive or down the scale and all the numbers are negative), then stick the sign on it at the end. That works because most problems ask "what is the heat loss" or "what is the heat gain" and we can get by with just a number and no sign. We answer heat loss is 64 kJ or heat gain is 64 kJ. That way no sign is required. (It doesn't make sense, for example to say it lost -64kJ or it gained +64 kJ). That's really say it lost lost or it gained gained.

• Chemistry - ,

Oh wow! Thank you!

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