Hi, can you guys please help me with this pysics question. I'm having too much trouble.
A 500 watt electric heater is used to boil 2kg of water at 40 degrees celcius into steam at 130 degrees celcius. <---assume constant volume.
a) Determine the amount of thermal energy required.
b) If the process takes 3.5 hours determine the efficiency of the heater.
Please somebody help me. I'm get headaches trying to figure it out.
2006-07-20 06:34:55 · 6 answers · asked by A 2 in Science & Mathematics ➔ Physics
I think you can do it easily.
2006-07-20 18:02:54 · answer #1 · answered by Sherlock Holmes 6 · 3⤊ 2⤋
From my hubby: He says there's an error in the question. Water actually turns to steam at 100 degrees celcius, not 130. Hope you find your answer.
2006-07-20 13:41:50 · answer #2 · answered by ~*Lady Beth*~ 4 · 0⤊ 0⤋
To raise the water from 40 C to 100 C takes 60 calories per gram. To evaporate the water takes an additional 540 calories per gram. To superheat the steam fo 130 degrees C takes an additional 15 calories per gram (if I have correctly remembered the specific heat of steam -- you should look it up.) Add all this up and multiply by 2000 grams to get the total calories required.
The electric energy consumed is 1750 watt-hours. Multiply this by the appropriate conversion factor to convert to calories. (Look this one up yourself.) Compare the number of calories consumed to the number required, and you're done.
2006-07-20 13:45:25 · answer #3 · answered by Anonymous · 0⤊ 0⤋
Sp.heat capacity of water is 1 cal/g
latent heat of vapourisation =540cal/g
Sp heat capacity of steam =0.48 cal/g
formula used heat=mst cal for heating
and mL for change of phase
heat for raising the temp of water
from 40 to 100=120000 cal.................(1)
heat for boiling 2 kg of water
and converting to steam=1080000cal (2)
heat for raising the temp of steam
from 100 to 130=28800 cal (3)
total heat=1228800 cal adding (1),(2)and(3)
=5141300j
so thermal energy required=5141300 joules
power=500j/s and time=3.5*3600 s
so heat dispensed=6300000joules
and efficiency=5141300/6300000
=0.816=>81.6%
2006-07-20 14:16:46 · answer #4 · answered by raj 7 · 0⤊ 0⤋
I don't know the answer to that question off the top of my head. However, I am sure that there is a formula in your Physics book that will help you solve that question.
2006-07-20 13:41:34 · answer #5 · answered by anonymous 2 · 0⤊ 0⤋
you have not given the values of specific heat of water, latent heat of vapourization pf water, specific heat of steam.
2006-07-20 13:40:25 · answer #6 · answered by shyam 2 · 0⤊ 0⤋