Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Introductory Physics Homework Help
Thermodynamics relating to temperature and phase changes V.2
Reply to thread
Message
[QUOTE="Mnemonic, post: 5223852, member: 557770"] [h2]Homework Statement [/h2] Suppose a room with 75 m3 of air also contains 80 kg of glycerol and the initial temperature in the morning is 16 °C. If 1.2 kWh of heat is added to the room between morning and afternoon, calculate the final temperature of the air in the room in the afternoon. Use 18 °C for the melting temperature, 200 J/kg for the specific latent heat of fusion, and 2400 J/kgK for the liquid and solid state specific heat of glycerol. Use 1000 J/kgK for the specific heat capacity and 1.2 kg/m3 for the density of air. Assume that the air and glycerol are at the same temperature as each other (i.e. they are isothermal with one another). Ignore the thermal mass of other materials in the room. (b) Now suppose that the glycerol was not present in the room. Calculate the final temperature of the air in this situation. [h2]Homework Equations[/h2] PV=nRT Q=mcΔt=ncΔt [h2]The Attempt at a Solution[/h2] Q=mcΔt Assuming no phase change: 1.2kWh=80*2400*Δt 1.2kWh=4320000J Δt=4320000/(2400*80) Δt=22.5C To bring temperature to melting point of glycerol Q=80*2400*2 Q=384000J Remaining Energy=4320000-384000 =3936000 Energy used in changing glycerol to sold to liquid: Q=ml =80*200 =16000J Therefore energy remaining is 3936000-16000 =3920000J Now to heat the air: Q=mcΔt Δt=3920000/(75*1.2*1000) =43.56 degrees Celsius Therefore final temperature of room is 16+43.56=59.65C This seems way too high. What did I do wrong? b) Temperature in room with no gylcerol: Δt=4320000/(75*1.2*1000) =48 Final Temp=64C [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Introductory Physics Homework Help
Thermodynamics relating to temperature and phase changes V.2
Back
Top