How Much Heat Is Needed to Melt Ice and Warm It to 30°C?

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Discussion Overview

The discussion revolves around calculating the heat required to raise the temperature of ice to a specified temperature, specifically how much heat is needed to melt ice and then warm it to 30°C. The context includes a problem-solving scenario related to thermodynamics, focusing on specific heat and latent heat concepts.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant calculates the total heat required as the sum of the heat needed to melt the ice and the heat needed to raise the temperature of the resulting water, arriving at 5500 cal.
  • Another participant suggests that the book's answer only accounts for the heat required to raise the temperature of water, not considering the melting of ice, which leads to a disagreement about the completeness of the book's solution.
  • A later reply speculates that the book's omission of latent heat is likely an oversight, reinforcing the idea that the exercise should consider both melting and heating.

Areas of Agreement / Disagreement

Participants express disagreement regarding the book's answer, with some believing it fails to account for the latent heat of fusion necessary to convert ice to water. There is no consensus on the correct approach, as differing interpretations of the problem persist.

Contextual Notes

Participants note that the problem involves latent heat, which is not addressed in the book's answer, indicating a potential limitation in the problem's framing.

Fantini
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Here's the problem.

How much heat is required to raise the temperature of $50.0$g of H$_2$O ice at $0.00^{\circ}$C to $30.0^{\circ}$C? Assume an average $1.00 \text{ cal/g}^{\circ}$C specific heat for water in this temperature range.

Since he said ice I assumed all $50.0$g is ice and therefore you need to both convert the ice to water than then heat the water. This amounts to

$$Q = mL + mc \Delta T = 50 \cdot 80 + 50 \cdot 1 \cdot 30 = 5500 \text{ cal}.$$

However the book answer is $1500$ cal, which I disagree. This is the amount of heat necessary to heat the water, but not to turn all ice to water.

Am I wrong?
 
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Well, the book is not converting the ice to water, right? If you just do the $m c \Delta T$ bit, you get the book's answer.

I would definitely have approached it the way you did, because I'm not sure I can conceive of a block of ice at $30^{\circ}\text{C}$. I suppose you could prove the book wrong if, in a vacuum (best-case scenario), heating the block of ice to $30^{\circ}\text{C}$ would have to melt it. It is true that the heat required to melt is considerably more than the heat required to raise the temperature. Still, my intuition is strongly on your side.
 
I'm guessing this is just an oversight from the book. He even labeled the exercise as Latent Heat. These things happen. :)
 
Fantini said:
I'm guessing this is just an oversight from the book. He even labeled the exercise as Latent Heat. These things happen. :)

So it looks like an example what would happen if you do not take latent heat into account. (Wink)
 

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