Thermodynamics relating to temperature and phase changes V.2

  • #1
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Homework Statement


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.

Homework Equations


PV=nRT
Q=mcΔt=ncΔt

The Attempt at a Solution


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
 

Answers and Replies

  • #2
Bystander
Science Advisor
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air and glycerol are at the same temperature as each other

To bring temperature to melting point of glycerol
Good.
glycerol to sold to liquid:
Good.
Forget something?
 

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