Question

A water vapor mass of mv at (130 c) was added to 0.2 kg of water inside a glass cup of mass 0.1 kg
which caused the mixture temperature to increase from 20 c to 50 c . Calculate the mass of the added water vapos.

water specific heat is 4186 j/kg.k
water vapor sepcific heat is 2010 j/kg.k
glass specific heat is 837 j/kg.k
latent energy of vaporization is 2.26 * 10^6 j/kg

Answer 1

@Michele_Laino

Answer 2

I think that we have an energy transfer, from vapor to glass and water, furthermore, I make the hypothesis that no condensation happens

Answer 3

I think there will be condensation as the water vapor is at 130c , and the mixture reached stability at 50 c (water condense at 100 c)

Answer 4

furthermore, we can say that water and glass are in thermal equilibrium before and during the energy transfer

Answer 5

what you say depends on the amounts of energies, not only on the temperatures

Answer 6

if we have a very large mass of vapor at 130 degrees, then we need of a very large heat decreasing, in order to get a condensation of such vapor

Answer 7

Yeah, that's for the other components but for the water vapor it's the source of the energy.
and no water vapor under 100

Answer 8

please wait a moment, since I have to answer to my phone...

Answer 9

Take, your time.

Answer 10

here I am

Answer 11

let's try the solution with condensation

Answer 12

in that case the system glass+ water receives an amount of heat. Such amount of heat comes from the condensation, plus a temperature decreasing of water vapor

Answer 13

Can we say that the total sum of all the energies (gained + lost) = 0 ?

Answer 14

yes! Nevertheless, we can write an equation which is in the subsequent form:
energy lost by water vapor = energy received by glass + water system

Answer 15

here is my equation:
\[\Large {\lambda _v}{m_v} + {m_v}{c_v}\left( {130 - 100} \right) = \left( {{m_g}{c_g} + {m_w}{c_w}} \right)\left( {50 - 20} \right)\]

Answer 16

the left side is the energy lost by water vapor, whereas the right side is the energy received by the system glass+water

Answer 17

please look at my equation above

Answer 18

after a simplification, i get:
\[\Large \begin{gathered}
\left\{ {{\lambda _v} + {c_v}\left( {130 - 100} \right)} \right\}{m_v} = \left( {{m_g}{c_g} + {m_w}{c_w}} \right)\left( {50 - 20} \right) \hfill \\
\hfill \\
{m_v} = \frac{{\left( {{m_g}{c_g} + {m_w}{c_w}} \right)\left( {50 - 20} \right)}}{{{\lambda _v} + {c_v}\left( {130 - 100} \right)}} \hfill \\
\end{gathered} \]

Answer 19

Thanks michele, I got it xD

Answer 20

When I tried to do it as you did, Gained = lost
I subtracted (100 -130) which was illogical

Answer 21

So the one which loses the energy (Old temprature - new temprature) ?

Answer 22

I got \(m_v=1.08\,grams\)

Answer 23

when an object loses heat, then we have: initial temperature - final temperature, and we have to write an equation like an balance of energy, namely
\(energy\; lost= energy \; received\)
Whereas, if you want to write initial temperature - final temperature for bodies which lose energy and final temperature - initial temperature for bodies which gains energy then you have to write an equation like a sommation:
\[\Large \sum\limits_i {{Q_i} = 0} \]
where index \(i\) runs over all bodies which participate to change energy.

Answer 24

I understand it now. This part used to confuse me.

Answer 25

ok! :)
If you need help again about such question and for other questions, of course, please tag, since I like very much Physics

Answer 26

Thanks again! If you found the answer for the previous question , explain it and tag me.

Answer 27

ok! :)