How Does an Electric Heater Affect the Melting Process of Ice?

AI Thread Summary
The discussion focuses on the heating and melting process of a 0.25kg piece of ice using an electric heater. Participants calculate the power output of the heater and the time required to raise the ice to its melting point and then to melt it completely. The correct calculations involve using the specific heat capacity of ice and the latent heat of fusion, leading to a power output of approximately 70W and a melting time of around 1178.6 seconds. Clarifications about the specific heat capacity of ice versus liquid water are also addressed. Overall, the thread emphasizes the importance of accurate temperature change calculations and understanding phase changes in thermal physics.
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1. A 0.25kg piece of ice at -30°C is warmed by an electric heater and the following graph of the temperature is produced (I just redrew the graph). Assume that there has been no loss of energy to the surroundings.
a) use the information on the graph to determine the power output of the heater.
b) explain how long it will take to get the ice to the melting point 0°C.
c) explain how long it will take to melt completely once the ice reaches 0°C. http://tinypic.com/r/5n26bk/8




2. Q=m(c)(Δt). P=Q/t. Q=m(L). L is Latent heat of fusion, m is mass, Δt is change in temperature, t is time, Q is quantity of heat, P is power and c isspecific heat capacity.



3. So for a) I did, Q=m(c)(Δt). Q=0.25(2100)(-30-(-10)). Q=-10500. Now that I have Q, I can find the power. P=Q/t. P=-10500/150. P=-70W. Is this the correct power output of the heater? How can it be negative? Then for b) I assumed that if it took 150s for the ice to get 20 degrees warmer, it would only take 75s to get 10 degrees warmer, no? For c). To get the time it takes the ice to completely melt I did, Q=m(L). Q=0.25(330000). Q=82500. Actually for c) I have no idea what I'm doing. Scratch that, the entire question is super confusing. I would appreciate it if someone could really break this down for me as physics is definitely not my strongest subject. Thank you.
 
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adjacent said:
Please fix your picture

How do I fix it?
 
yesgirl10 said:
How do I fix it?
You can't edit the post now.So go to the reply box,click on advanced editor button

Click on the attachment button and attach it.
 
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yesgirl10 said:
a)

I did, Q=m(c)(Δt). Q=0.25(2100)(-30-(-10)). Q=-10500. Now that I have Q, I can find the power. P=Q/t. P=-10500/150. P=-70W. Is this the correct power output of the heater? How can it be negative?

Let's go one by one.
What is ##\Delta t##?
Is it final temp-initial temp or initial temp -final temp?
 
adjacent said:
Let's go one by one.
What is ##\Delta t##?
Is it final temp-initial temp or initial temp -final temp?

I think it's the final temp.
 
yesgirl10 said:
I think it's the final temp.

That was not one of the options
 
Last edited:
dauto said:
That was not one of the options

Oh, woops. Well Δt=t2-t1, so final temp-initial temp I believe. So does that mean that Δt is actually -10-(-30)?
 
  • #10
Is that correct?
 
  • #11
Yes. Done with a and b. You can update the graph. How does it continue after 225 seconds ?
I think you are doing just fine!
Now, in c, being at the melting point means that two phases co-exist. Any heat you put in melts ice. Your heater gives off 70 W and you need 0.25 kg * 333.55 kJ/kg = 83 kJ to melt the block. How long does it take ?

(I was surprised, too. A long time ago, though...)
 
  • #12
BvU said:
Yes. Done with a and b. You can update the graph. How does it continue after 225 seconds ?
I think you are doing just fine!
Now, in c, being at the melting point means that two phases co-exist. Any heat you put in melts ice. Your heater gives off 70 W and you need 0.25 kg * 333.55 kJ/kg = 83 kJ to melt the block. How long does it take ?

(I was surprised, too. A long time ago, though...)

Thanks for the reply!
Ohh ok this is making sense, so since Q=83500J and the heater is giving out 70W or 70J/s I did 83500/70 which gave me approximately 1178.6 seconds for the ice cube to completely melt. Right?
 
  • #13
Very much so. Was the 330 kJ/kg a given? (I google 333.55)
 
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  • #14
BvU said:
Very much so. Was the 330 kJ/kg a given? (I google 333.55)

Thank you so much for your help! :) And yes that is what they gave me.
Thanks again!
 
  • #15
Sorry to resurrect this post, but I am doing the same question and am wondering about the formulas used... why is the value of c = 2100? Isn't the heat capacity of water 4200 J?
 
  • #16
LonelyElectron said:
Sorry to resurrect this post, but I am doing the same question and am wondering about the formulas used... why is the value of c = 2100? Isn't the heat capacity of water 4200 J?
Cice is not the same as Cliquid for water.
 
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  • #17
gneill said:
Cice is not the same as Cliquid for water.

I CANT BELIEVE I MISSED THIS. YOURE BRILLIANT.
 

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