[Blueberries116]

The graph from below shows Temperature with respect of heat in the thermal interaction between a sample of [itex]400\,g[/itex] of liquid water and another sample of [itex]200\,g[/itex] of ice. What is the final composition of the mixture?



The alternatives given are as follows:

[itex]\begin{array}{ll}
1.&\textrm{400 g. of liquid water and 200 g. of ice}\\
2.&\textrm{400 g. of liquid water and 25 g. of ice}\\
3.&\textrm{600 g. of ice}\\
4.&\textrm{600 g. of liquid water}\\
5.&\textrm{575 g. of liquid water and 25 g. of ice}\\
\end{array}[/itex]

I'm not sure exactly how can I get the composition of the mixture. I'm assuming that the horizontal line from which is not given any information of the heat represents the transition between the solid and liquid phase. But how can the composition of the mixture be found?. Can somebody help me here please?.

[AWK]

The answer can be seen in the sketch

[Borek]

I don't like the plot, it is confusing as hell. Apparently it can be read as "when Q of water goes up, temperature of water goes down" which is a pure nonsense. Perhaps it is intended to mean "temperature vs amount of heat transferred", but then it should be clearly described as that.

Upper line seems to be describing the water - its temperature goes down. Lower line is about the ice - it is initially heating up, then melting. Unfortunately, if that's the case IMHO there is no way to say which of the answers given is a correct one. It is trivial to eliminate some of them, but all we can say is that at the moment water temperature went to zero system got to equilibrium, I don't think it is possible to tell if all the ice was melted or not.

[Borek]

The answer can be seen in the sketch

Perfect example of an answer that is as unhelpful as possible

[AWK]

When you answer the tests, you don't have much time to count. You can reject answers 1, 2 and 3 right away by looking at the sketch and the figures. Why?
 The heat needed to cool the water completely to 0°C equals to 400·40·1 = 16000 cal. The heat needed to completely melt the ice equals to 200·80 = 16000 cal. So there is no heat needed to heat the ice, which means the only possible answer is ...
All can be done even without a calculator.

For calculations, the heat released during water cooling must be split into heat needed for warming of ice to 0°C and heat needed to melt some ice. This way you can check the selected answer (200·20·0.5+175·80=... which is the correct answer)

[Blueberries116]

I don't like the plot, it is confusing as hell. Apparently it can be read as "when Q of water goes up, temperature of water goes down" which is a pure nonsense. Perhaps it is intended to mean "temperature vs amount of heat transferred", but then it should be clearly described as that.

Upper line seems to be describing the water - its temperature goes down. Lower line is about the ice - it is initially heating up, then melting. Unfortunately, if that's the case IMHO there is no way to say which of the answers given is a correct one. It is trivial to eliminate some of them, but all we can say is that at the moment water temperature went to zero system got to equilibrium, I don't think it is possible to tell if all the ice was melted or not.

I'm glad to know to I'm not the only one who found this graph confusing. But essentially what it bothers me is that no heat is given between the transition of solid phase to liquid.

I think the intended meaning is that the upper portion of the graph which has a negative slope is referring to the liquid water which is cooling to be solid as the ice warms up and melts but some of ice remains. Then the question remains, how to calculate this. Supposedly the answer to this question is [itex]\textrm{575 g. of liquid water and 25 g. of ice}[/itex] according to my answers sheet, but I have no idea how to justify this.

[Blueberries116]

When you answer the tests, you don't have much time to count. You can reject answers 1, 2 and 3 right away by looking at the sketch and the figures. Why?
 The heat needed to cool the water completely to 0°C equals to 400·40·1 = 16000 cal. The heat needed to completely melt the ice equals to 200·80 = 16000 cal. So there is no heat needed to heat the ice, which means the only possible answer is ...
All can be done even without a calculator.

For calculations, the heat released during water cooling must be split into heat needed for warming of ice to 0°C and heat needed to melt some ice. This way you can check the selected answer (200·20·0.5+175·80=... which is the correct answer)

The answer to this question is [itex]575\,g[/itex] of liquid water and [itex]25\,g[/itex] of ice. How can you obtain those from the calculation you propose?. I can't find how to get to that answer from your last sentence.

[Borek]

Supposedly the answer to this question is [itex]\textrm{575 g. of liquid water and 25 g. of ice}[/itex] according to my answers sheet, but I have no idea how to justify this.

By calculations, ignoring the plot.

I told you already twice: split the job into single steps and do calculations stepwise, don't rush ahead.

Has the water enough heat to heat up the ice to 0°C?

Has the water enough heat to further melt all of the ice?

[AWK]

200·20·0.5+175·80=...

200·20·0.5+x·80 = 16000 cal where x is a mass of ice melted.