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Topics: Liquid, Water, Temperature Pages: 2 (445 words) Published: February 7, 2013
Analysis

The melting temperature of sodium thiosulfate can be determined by checking our data. Since a substances temperature remains constant while it undergoes a phase change, we can see that our sample changed from a sold to a liquid at 46.0°C. This is the temperature of the sample between time periods t1= 130s and t2= 480s.

In order to find the Heat of Fusion (∆H Fusion) of sodium thiosulfate we used to calculate how much energy the water bath lost during the sample’s melting period. That energy is what melted the sample. The water was a liquid and we can find out how much energy it lost if we know the number of moles of water (n[pic]), the heat capacity of water (Cp[pic]) and the change in temperature of the water throughout the samples melting period. If we know those values, then we can place them into the equation HEAT=nCp∆T.

The number of moles of water in the bath were:

[pic][pic]

and the heat capacity of liquid water is [pic]. The data table shows that our water went from T130= 62.9°C= 335.9°k down to T450= 54.5°C= 327.5°k while the sample melted. This is a change in temperature of [pic].

Putting in these values into the equation for heat transfer we get the following amount of energy that was absorbed by the sample while it melted:

[pic]

To find the ∆H Fusion we need to take this absorbed energy and divide it by the number of moles that were in our sample. Our sample had a mass of 15.12g, which is equal to the following number of moles:

[pic]

The Heat of Fusion is found by dividing the energy it took to melt the sample by the number of moles in our sample:

[pic]

To find the heat capacity of the sodium thiosulfate in the liquid state we need to choose a time interval to work with. Choosing the temperature values between T1= 390s and T2= 460s, we see that the water temperature went from 46.7°C(319.7°k) to 47.3°C(320.3°k), which corresponds to an increase of ∆T sample= 320.3°k-319.7°k = 0.6°k.

Finding the...