# Heat Transfer Help: Find Time to Melt Ice

• -Dragoon-
In summary, the student correctly used equations to solve for the energy and power needed to heat the ice and melt it. However, there were some significant rounding errors in the calculations which resulted in a slightly incorrect final answer. It was suggested to be more careful with significant figures.

## Homework Statement

A 0.25 kg piece of ice is warmed by an electric heater. Assume that there has been no loss of energy to the surroundings. It takes 150 seconds to heat the ice from -30° to -10°. How much additional time after 150 seconds will be required to melt all of the ice, assuming that the power of the heater is constant?

I made this on my computer to help me visualize the problem:
http://img824.imageshack.us/i/energytransferphysics.jpg/

## Homework Equations

Qwarm ice = miceciceΔtice
Q melt ice = miceLfusion
P = ∆Q/∆t

## The Attempt at a Solution

Firstly, I use the first equation to find the energy it takes to heat the ice to the melting -30° to -10° which yields about 10500 J. I know this takes a 150 seconds, so I use the third equation to find the power which gives a value of 70 W. Now that I have the power, I can find the time it takes to heat the ice from -10° to 0° using the first equation. I use the first equation again to get a value of 5250 J and divide this by 70 W, which is 75 seconds to heat from -10° to 0°. Finally, I use the second equation to find the energy needed to melt the ice and gives me a value of 82,500 J. I divide this by the power (70W) to find the energy and finally yield a value of 1100 S. 1100+75 = 1175 seconds. After the 150 seconds it took to heat the ice from -30° to -10°, it will take another 1175 seconds for the ice to heat from -10° to 0° and melt.

Did I do this correctly?

Retribution said:
Firstly, I use the first equation to find the energy it takes to heat the ice to the melting -30° to -10° which yields about 10500 J. I know this takes a 150 seconds, so I use the third equation to find the power which gives a value of 70 W. Now that I have the power, I can find the time it takes to heat the ice from -10° to 0° using the first equation. I use the first equation again to get a value of 5250 J and divide this by 70 W, which is 75 seconds to heat from -10° to 0°. Finally, I use the second equation to find the energy needed to melt the ice and gives me a value of 82,500 J. I divide this by the power (70W) to find the energy and finally yield a value of 1100 S. 1100+75 = 1175 seconds. After the 150 seconds it took to heat the ice from -30° to -10°, it will take another 1175 seconds for the ice to heat from -10° to 0° and melt.

Did I do this correctly?
Your general approach seems perfectly fine to me.

But I think you're making some significant rounding/truncation errors. For example, you've implied that
82,500/70 → 1100.​
That's significantly different that what I would say, which would be closer to something around 1179.

Anyway, just be careful with your significant figures.

collinsmark said:
Your general approach seems perfectly fine to me.

But I think you're making some significant rounding/truncation errors. For example, you've implied that
82,500/70 → 1100.​
That's significantly different that what I would say, which would be closer to something around 1179.

Anyway, just be careful with your significant figures.

Ah, I see where I made the mistake.

Thanks for the help!