How much heat ΔQ is consumed to raise the ice temperature

  • Thread starter Firben
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  • #1
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1. How much heat ΔQ is consumed to raise the ice temperature from Ti = -20 Celsius to meltingpoint Tf = 0 ? how long does it take ?



2. dQ/dt = H = 300 W

dQ = cmΔT




3 I got it to be 2.49 * 10^-21 J. Is it right ?
 

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  • #2
Andrew Mason
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1. How much heat ΔQ is consumed to raise the ice temperature from Ti = -20 Celsius to meltingpoint Tf = 0 ? how long does it take ?



2. dQ/dt = H = 300 W

dQ = cmΔT




3 I got it to be 2.49 * 10^-21 J. Is it right ?
You will have to provide a bit more information. How much ice are we talking about? Does the heat source provide 300 W?

AM
 
  • #3
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I dont know anything about the mass.

c = 1/m * dQ/dt (1)

c = 4190 kJ/kg (heat capacity of water)

Q = mcΔT (2)

From equation (1): m = (dQ/dt)/c (3)

m = 300W/4190 k//kg = 0.0715 kg

(2) Q = (0.0715 kg)(4190 kJ/kg)(273-15-253.15)K <==>

Q = 5.99 =~ 6 J

t = Q/H (4) (heat/power)

t(time) = 6.0 J/300 W = 0.02 J/s = 2 min

Is it right ?
 
  • #4
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How did you get the equation (1)?
 
  • #5
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I just solved it, an oversight
 
  • #6
Andrew Mason
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The units are wrong. c has units of energy/mass. 1/m(dQ/dt) has units of energy/(mass x time).

Your derivation of m makes no sense. You are using the wrong heat capacity. You have to use the heat capacity of ice, not liquid water. Are you sure you got it solved?

AM
 

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