What is the final temperature of water if 175 g steam at 150 C loses 423 KJ of energy?
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:
- Going from steam to water --> Q = m * heat of vap
- 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?