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posted by  Ruger_Mk2 on 5/7/2008 5:34:07 PM  |  status: Closed  

HELP ASAP!!!

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posted by AliAli on 5/7/2008 10:30:38 PM  |  status: Live
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Ruger_Mk2's comment:
"Thank you"
Response Details:
Heat is transferred with 100% efficiency from the hot brass object to water as its dropped in.  We can stylize it as follows:

Qg = Ql where Qg is heat gained, Ql is heat lost

Qg is for the water and the formula is mcΔT where m is mass, c is specific heat capacity, and dT is the temperature difference & should always be a positive value.  A similar equation can be written for brass

mw           cw      (6.8 - 4.2)  =     mb cb (96.6 - 4.2)  Now plug in what you know

2.73kg (4186) (6.8 - 4.2) = (.64 kg) cb (96.6 - 4.2) 

NOw just solve for cb, the specific heat capacity of the brass.
Best of Luck & please rate.
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posted by richmondMike on 5/7/2008 10:42:36 PM  |  status: Live
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Ruger_Mk2's comment:
"Thank you"
Response Details:
Mass of the water is mwater = 2.73kg.
The specific heat of water is cwater = 4186J/kg 0C.
ΔTwater = 6.80C - 4.20C = 2.60C
...(1).
Mass of the brass is mbrass = 0.64kg.
The specific heat of brass cbrass.
ΔTbrass = 6.80C  - 96.60C = -89.80C
...(2).
Let (1) = (2) and solve for cbrass = 517J/kg 0C.
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posted by Kayen on 5/8/2008 2:59:20 AM  |  status: Live
Asker's Rating: Helpful   
Ruger_Mk2's comment:
"THANKS!!!!!"
Response Details:
Heat lost by brass Q1 = m1c1Δt1
heat gained by water, Q2 = m2c2Δt2
  Q1 = Q2
    
   0.64.c2(96.6-6.8) = 2.73x 4186x(6.8-4.2)
Hence,
     c2 = 29712/ 57.47 = 517 J/kg.oC
 
 
Kayen (Enjoys helping you uderstand..)
Please do not forget to Rate my Response.
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