- #1
PHYclueless
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Hello all. I have two questions I need help with but I'll post them in separate posts. I'll attach the problem and the work I've attempted. If someone could give me another idea I'd appreciate it. Thank you!
To make steam, you add 5.6x10^5J of heat to .220kg of water at an initial temperature of 50c. Find the final temperature of the steam, assuming a specific heat for steam = 2010J/(kg K), latent heat of fusion for water=33.5x10^4J/kg, and a latent heat of vaporization =22.6x10^5J/kg.
This is what I started working on and my answer comes out to be 38 degrees but the true answer is 138. Not sure where I'm going wrong here.
mass(specific heat)(change in temp)+mass(latent fusion)+mass(specific heat)(Tf-Ti)+mass(latent vaporization)=Energy
.220kg(4186J/kgc)(50c)+.220kg(33.5x10^4J/kg)+.220kg(4186J/kgc)(Tf-100c)+.220kg(22.6x10^5)=5.6x10^5J
Thanks!
To make steam, you add 5.6x10^5J of heat to .220kg of water at an initial temperature of 50c. Find the final temperature of the steam, assuming a specific heat for steam = 2010J/(kg K), latent heat of fusion for water=33.5x10^4J/kg, and a latent heat of vaporization =22.6x10^5J/kg.
This is what I started working on and my answer comes out to be 38 degrees but the true answer is 138. Not sure where I'm going wrong here.
mass(specific heat)(change in temp)+mass(latent fusion)+mass(specific heat)(Tf-Ti)+mass(latent vaporization)=Energy
.220kg(4186J/kgc)(50c)+.220kg(33.5x10^4J/kg)+.220kg(4186J/kgc)(Tf-100c)+.220kg(22.6x10^5)=5.6x10^5J
Thanks!