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Would anyone be able to help me calculate the following questions? I have the answers, I just want to understand the process. If you are able to give me any feedback, would you be able to write your answer in simple steps plugging in the values? Thank you.

I'm not demanding answers and if it seems that way, I'm sorry! I really do just want to understand HOW to get to the answers (no matter what values I'd plug into the equation).

Steam at 100 °C is mixed with 191.2 g of ice at – 44.1 °C, in a thermally insulated container, to produce water at 53.2 °C. Ignore any heat absorption by the container.

How much energy is required to bring all the ice up to 0 °C? ANSWER: 17600.0 J

How much energy is required to melt the ice into water at 0 °C? ANSWER: 63700.0 J

How much energy is required to raise the temperature of this melted water to its final temperature (53.2 °C).ANSWER: 42600.0 J

How much energy must then be supplied by the steam to change the state of 191.2 g of ice at –44.1 °C to water at 53.2 °C? ANSWER: 124000.0 J

What is the final mass of water in the cup at 53.2 °C? ANSWER: 242.0 g

Ali
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Alexis Maria
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    I notice that four users have voted to close this question with the reason: "Homework questions must seek to understand the concepts being taught, not just demand a solution. For help writing a good homework question, see: How to ask a homework question?." However, based on the question, it seems to me that @AlexisMaria does wish to understand the concepts being taught (e.g., read the first paragraph). The OP is not demanding a solution because the solutions are already provided; rather, the OP would like to understand the process behind the answers ... – Amitesh Datta Aug 25 '13 at 05:22
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    ... So, although I agree this question is borderline, I think this is more so due to the phrasing of the question. The intent of the question doesn't seem to be simply demanding a solution and thus I don't think a close vote is warranted. (Also, see the comment below by the OP; I am quite confident that s/he is sincere and I do hope that this question isn't closed.) – Amitesh Datta Aug 25 '13 at 05:23
  • Thank you guys. I'm not demanding answers and if it seems that way, I'm sorry! I really do just want to understand HOW to get to the answers (no matter what values I'd plug into the equation). – Alexis Maria Aug 25 '13 at 05:26
  • Alexis Maria, could you please edit your answer and add the comment you just posted at the very top? I think it's best to draw attention to the fact that you do want to understand the concepts being taught in order to prevent your question from being closed. (If the question is closed, then please don't take it personally! You can always ask further questions and/or refine this question.) – Amitesh Datta Aug 25 '13 at 05:32

1 Answers1

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  1. The amount of energy required to change the temperature of a substance of: mass $ M $, specific heat $S$ is given by $$\text{Energy} = MS\Delta T$$ where $\Delta T$ is the change in temperature that the substance undergoes. This equation holds good only if the substance doesn't change its phase (solid to liquid or liquid to gas etc). In your first question, the ice is initially at $ -44.1^\circ $ C and then goes to $ 0^\circ $C (still being ice at $ 0^\circ $). You know the mass of the ice, you can find out the specific heat from some science data book (it should be mentioned in the question ideally). $$ $$ Here, the mass is given to be $ 191.2 $g and $\Delta T = 44.1^\circ = 44.1 K$. The specific heat of ice is $ 2.11 J \cdot g^{-1} \cdot K^{-1} $ So, the energy that must be supplied is $$ (191.2)\cdot(2.11)\cdot(44.1) \approx 17800 \, J$$ The difference in the answers is because the value of specific heat that the book-author has used is $\approx 2.08$ I used the one available on wikipedia.

  2. If a substance undergoes phase transition, (solid to liquid or liquid to gas or vice versa), then the amount of energy required/given out is given by the expression: $$ \text{Energy} = ML $$ where $L$ is the latent heat of fusion or vapourisation depending upon the process involved. That should be given in the question statement or in the data book. $$ $$ Here,the phase transition is from solid to liquid. So, we must use the latent heat of fusion. Wikipedia says that for water, the latent heat of fusion is $ 334 \text{ kJ/kg} = 334 \text{ J/g } $. So, using the equation, we get the energy required as: $$ (191.2) \cdot (334) \approx 63800 \,J $$

  3. Pretty much the same as the first part of the problem.

  4. Pretty much the same as the second part of the problem.

However, I'd like to know why you had to ask this - after all, all this information should've been explained nicely in your textbook - in much greater depth than what I have presented here, with examples and stuff. Did you give a read to the textbook? If you did and didn't understand, then that's fine. But I wouldn't appreciate not reading the text and straight away asking the question here. (I've done that before and since someone pointed out, I always research before asking).

Parth Thakkar
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  • Hmmm, yes I've read the text. I don't really understand your answer either as I've given you the question plus the values. – Alexis Maria Aug 25 '13 at 06:06
  • Ok. Showing the working for the first and the second problems. – Parth Thakkar Aug 25 '13 at 06:08
  • Edited. Does that help? – Parth Thakkar Aug 25 '13 at 06:22
  • Yes, it does. Thank you. Unfortunately, I need the values to be exactly the same as the answers I gave. I found the constants they gave me though: Cwater = 4186. J/(kg °C)

    Cice = 2090. J/(kg °C)

    Lf,water = 3.33 × 105 J/kg

    Lv,water = 2.26 × 106 J/kg

     – Alexis Maria Aug 25 '13 at 06:33
  • I bet you can do the calculations now. Just be careful with the units. Your $C_{ice}$ for instance is given in J/kg. Be sure to convert it into J/g as our weight is in grams ( pun very much intended :D ). And for calculations, I must say, you will almost never get what is called exact values, especially for problems that involve actual experimental data. You'll get nice numbers only if the question is set up such that you get a nice number. Make approximation your good friend and it'll be fun. – Parth Thakkar Aug 25 '13 at 06:40