Latent Heat Of Fusion

  • Thread starter STEMucator
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  • #1
STEMucator
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Homework Statement



How much heat is required to heat 0.2kg of ice from -20°C to 30°C?

Homework Equations



Total heat lost or gained : ##Q = mcΔt## where ##Δt## is the change in temperature and c is the heat capacity.

Latent Heat Fusion : ##Q = mL_f##

The Attempt at a Solution



From my understanding, I think I need to break this problem down into states.

The ice will start melting at 0°C ( since ice is just condensed water ). So the ice must be heated from -20°C to 0°C for it to melt.

Then we must use Latent Heat to melt the ice.

Finally we would have to heat the water from 0°C to 30°C.

So the amount of heat we would need can be expressed by the equation :

##Q_T = Q_{Ice} + Q_{Melt} + Q_{Water} = mc_{Ice}Δt_{Ice} + mL_f + mc_{Water}Δt_{Water}##

Then plugging and chugging I get :

##(0.2 kg)(2100 J/kg°C)(0°C + 20°C) + (0.2 kg)(330000 J/kg) + (0.2)(4200 J/kg°C)(30°C - 0°C) = 99600J##

Therefore 99600J of heat would be required to heat the ice from its original state to a heated state of water.

Does this look okay?
 

Answers and Replies

  • #2
TSny
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Looks good!
 
  • #3
Doc Al
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Yep, looks good to me as well.
 
  • #4
BruceW
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yep. looks good to me too. Although you have used 2 significant figures in your values, so does it make sense to have 3 significant figures in your answer?

edit: hehe, everyone seems to have posted a reply near-simultaneously.
 
  • #5
STEMucator
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BruceW said:
edit: hehe, everyone seems to have posted a reply near-simultaneously.

You're all so fast lol. Thanks for the confirm guys :).
 
  • #6
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"edit: hehe, everyone seems to have posted a reply near-simultaneously. " It's because it was an easy problem :-)
 

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