# why not!

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why not!
Part I:
Insert a complete data table, including appropriate significant figures and units, in the space below. Also include any observations you made over the course of Part I.
Part II:
Insert a complete data table, including appropriate significant figures and units, in the space below. Also include any observations you made over the course of Part II.
Calculations:
Show your work and write a short explanation with each calculation.
Part I:
Calculate the energy change (q) of the surroundings (water) using the enthalpy equation

qwater = m × c × ΔT.

We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL.

The water has absorbed the heat of the metal. So, qwater = qmetal

Using the formula qmetal = m × c × ΔT, calculate the specific heat of the metal. Use the data from your experiment for the metal in your calculation. Aluminum: q=26*4.18*6.3=684.684, c=q/(m*^T)=3.913 Zinc: q=26*4.18*10.3=1119.404, c=q/(m*^T)=2.608
Iron: q=26*4.18*7.8=847.704, c=q/(m*^T)=3.130
Copper: q=26*4.18*9.2=999.856, c=q/(m*^T)=2.608
Part II:
Calculate the energy change (q) of the surroundings (water) using the enthalpy equation

qwater = m × c × ΔT.

We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL. Show ALL your work.

The water has absorbed the heat of the metal. So, qwater = qunknown metal

Using the formula qunknown metal = m × c × ΔT, calculate the specific heat of the metal. Use the data from your experiment for the unknown metal in your calculation. Show ALL your work. A: water q=24.0*4.18*2.3=230.736, c=230.736/(15.262*2.30)=6.573 B: water q=24.0*4.18*6.9=692.208, c= 692.208/(25.605*6.9)=3.918 C: water q= 24.0*4.18*2.8=280.896, c=280.896/(20.484*2.8)=4.897

Conclusion:
Use the given specific heat capacity values below to calculate the percent error of the experimental specific heat capacity that you determined in

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