Problem 17.97 Part A If no heat is lost to the surroundings, what is the final t

Question

Problem 17.97 Part A If no heat is lost to the surroundings, what is the final temperature of the system? In a container of negligible mass, 4.10×10-2 kg of steam at 100 "C and atmospheric pressure is added to 0.210 kg of water at 51.0 CO CO T= 58.333 Submit My Answers Give Up Incorrect; Try Again; 5 attempts remaining Part B At the final temperature, how many kilograms are there of steam? m= kg Submit My Answers Give Up Part C How many kilograms are there of liquid water? m= kg Submit My Answers Give Up

Explanation / Answer

constants used:

latent heat of vaporisation=2257 kJ/kg

specific heat of water=4.187 kJ/(kg.K)

let final temperature be T degree celcius.

as no heat is lost to the surrounding,

then heat lost by the steam=heat gained by the water

==> heat generated by conversion of steam to water+ heat generated by decreasing temperature of converted water from 100 degree celcius to T

=heat required by 0.21 kg water to increase temperature from 51 degree celcius to T

==>4.1*0.01*2257+4.1*0.01*4.187*(100-T)=0.21*4.187*(T-51)

==>4.1*0.01*2257+4.1*0.01*4.187*100+0.21*4.187*51=T*(0.21*4.187+4.1*0.01*4.187)

==>T=147.0559 degree celcius

but it is not possible

so it means all of steam is not converted to water and the final temperature is 100 degree celcius.

part B:

let m kg of steam remains.

then 4.1*0.01-m kg of steam is converted from steam to water at 100 degree celcius and the heat generated is used to increase temperature of 0.21 kg water from 51 degree celcius to 100 degree celcius

==>(4.1*0.01-m)*2257=0.21*4.187*(100-51)

==>m=2.191*10^(-2) kg

hence 2.191*10^(-2) kg of steam remains

part C:
mass of liquid water=4.1*0.01-m+0.21

=0.229 kg

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