# Thermal energy question

1. Apr 30, 2014

### yesgirl10

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.

2. Apr 30, 2014

3. May 1, 2014

### yesgirl10

How do I fix it?

4. May 1, 2014

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.

5. May 4, 2014

### yesgirl10

Like this?

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6. May 4, 2014

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

7. May 4, 2014

### yesgirl10

I think it's the final temp.

8. May 4, 2014

### dauto

That was not one of the options

Last edited: May 4, 2014
9. May 4, 2014

### yesgirl10

Oh, woops. Well Δt=t2-t1, so final temp-initial temp I believe. So does that mean that Δt is actually -10-(-30)?

10. May 6, 2014

### yesgirl10

Is that correct?

11. May 6, 2014

### BvU

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. May 6, 2014

### yesgirl10

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. May 6, 2014

### BvU

Very much so. Was the 330 kJ/kg a given? (I google 333.55)

14. May 6, 2014

### yesgirl10

Thank you so much for your help! :) And yes that is what they gave me.
Thanks again!!