Heat specificity

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Data Table 2. First, Metal.
Objects
Mass (g)
Tinitial
Tfinal
DT
C (cal/g °C)
Water in
calorimeter
25
23
28
5
1.0000
First metal
17.3
98
28
-70
0.1032
Data Table 3. Second, Metal.
Objects
Mass (g)
Tinitial
Tfinal
DT
C (cal/g °C)
Water in
calorimeter
25
22
23
1
1.0000
Second metal
12.9
98
23
-75
0.0258
Analysis
The quantity of heat energy wasted by the metal is equal to the heat obtained by the calorimeter and water, plus heat lost to the atmosphere. Lost or won heat is:
Q = cmΔT
The calculation for measuring 1st metal:
Then for the basic heat measurement of the first metal in the first set of data:

CS = cwmwΔTw / msΔTs

CS= [(1.00 cal/g · °C) (25 mL) (5 °C)] / [(17.3 g) (70 °C)]

CS= 125 / 1211

CS= 0.1032 cal/g · °C

Calculation for 2nd metal calculation

So for calculating the specific heat of the first metal in the first set of data:

CS = cwmwΔTw / msΔTs

CS= [(1.00 cal/g · °C) (25 mL) (1 °C)] / [(12.9 g) (75 °C)]

CS= 25 / 967.5

CS= 0.0258 cal/g · °C

Conclusions

The comparison of the specific heat results obtained from the calculation using data results with those in the table of the lab manual. Metal in first experiment indicates that it can be steel while the second metal proves to be lead.

Questions

Why is it a good idea to start with room temperature water in the calorimeter?

Room temperature water negates the effects of the environment outside the calorimeter and act as control in the experiment for this practical.

Why did we ignore the calorimeter in our calculation, although it is listed in the original equation?

The calorimeter was ignored since it change in heat was negligible and has no effect in the results and final calculation.

When eating apple pie, you may have noticed that the filling seems to be much hotter than the crust. Why is this? What can you conclude about the specific heat of the fillings vs. the specific heat of the crust?

The crust is more densely packed than the crust. Energy is absorbed at a lower level at filling compared to crust. Finally, the filling becomes hotter than crust hence different in heat capacities that conforms the Dulong and Petit law.

Is the heat exchange between the metal and the water in the calorimeter by radiation, conduction, or convection? Why?

The exchange of heat between the metal and the water in the calorimeter was a result of conduction. The metal was heated by boiling water. The metal was removed out of the boiling water and placed into the calorimeter. The temperature rose by conduction process.

References

Kleiven, Andreas, and Helge Skarestad. “Specific Heat Capacity of Metals.” TFY4165 Termisk fysikk, Nttnu (2014).

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