Find the final temperature of water

  • Thread starter jwxie
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  • #1
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Homework Statement



What is the final temperature of water if 175 g steam at 150 C loses 423 KJ of energy?

Homework Equations


Q = M *C_p * Δ T
Q = m * heat of (phase_change)

The Attempt at a Solution



This is how I would go about solving this problem.

Going from steam to water requires energy of two types:
  1. Going from steam to water --> Q = m * heat of vap
  2. Cooling from 150C to 100C --> Q = m * Specific heat of steam * Δ T

Furthermore, we are told that steam is finally in the form of water, which has an unknown final temperature, so we have
Q = m * specific heat of water * ΔT

Add them together, and equates with the heat loss (which is given)

- 423KJ = Q (from steam to water) + Q (cooling from 150 to 100) + Q (cooling from 100 to unknown final temperature)

Is this correct?

If it is, my question would be: is it also correct to make Q (from steam to water) negative? I don't remember seeing heat of condensation. Can I make heat of evaporation negative in this case?

Anything else needs to be negative also?

Thank you.
 

Answers and Replies

  • #2
Redbelly98
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You are on the right track. All the Q's will be negative, since cooling to lower temperatures and condensing form vapor to liquid all involve loss of energy.
 
  • #3
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You are on the right track. All the Q's will be negative, since cooling to lower temperatures and condensing form vapor to liquid all involve loss of energy.

Thanks. But if all Qs are negative, unknown delta T is x = 13.6575
It is still above the boiling point though.
I thought it was safe to assume that it is no longer steam, but water, fully in liquid form.\\

Here is my setup
-423000 = (-2260 * 175g) + (-2.0 * 50 *175) + (-4.184 * 175 * x)
x = delta T = 13.6575
Hence, 150 - 13.6575.

THanks.
 
  • #4
Borek
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I thought it was safe to assume that it is no longer steam

Apparently it was not safe :tongue:

-423000 = (-2260 * 175g) + (-2.0 * 50 *175) + (-4.184 * 175 * x)

I wanted to check the result, but this is as cryptic as it can be - no units, no explanation of what is what. Sorry, I am not going to take off my tin foil hat to read your mind, it is too noisy out there.
 
  • #5
Redbelly98
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Thanks. But if all Qs are negative, unknown delta T is x = 13.6575
It is still above the boiling point though.
I thought it was safe to assume that it is no longer steam, but water, fully in liquid form.\\

Here is my setup
-423000 = (-2260 * 175g) + (-2.0 * 50 *175) + (-4.184 * 175 * x)
x = delta T = 13.6575
Hence, 150 - 13.6575.

THanks.
Looks about right, except for one detail: earlier you (correctly) wrote:
- 423KJ = Q (from steam to water) + Q (cooling from 150 to 100) + Q (cooling from 100 to unknown final temperature)
So x is the temperature change after the water has reached 100 C.

By the way, Borek has a good point. Including units would be helpful and depending on who is grading your assignment -- or an exam you take in the future -- you might lose points by just writing down numbers without units.
 

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