6.03 Calorimetry Essay

1569 Words May 20th, 2013 7 Pages
06.03 Calorimetry: Lab Report
Before You Begin: You may either copy and paste this document into a word processing program of your choice or print this page.
Instructions: This is a two-part lab. Be sure to follow all steps given in the lab and complete all sections of the lab report before submitting to your instructor.
Procedure:
Part I: Determining the Specific Heat of a Known Metal 1. Place a plastic measuring trough on top of the digital balance, and press the "tare/on" button so that the mass of the trough will be "ignored." The digital balance should read "0.000 g" after this step. 2. Measure out 25 to 45 grams of one of the four metals on the trough. This may require several "clicks." 3. Record the mass of the
Use the data from your experiment for the metal in your calculation. q(water) = - q(metal) q(metal) = - 1478 Joules q(metal) = m × c × ΔT m = 27.776 g
ΔT = T(mix) - T(metal)
ΔT = 38.9 - 100.5 = - 61.6
C = q(metal) / m x ΔT
C = -1478 / (-61.6 x 27.776 )
C = 0.864 J / (g × °C)

Part II: 1. 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
(27.776 g) * ( 4.18 J/g *C ) * ( 25.3 C ) Q=2944kJ

2. Using the formula q 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.

qmetal = -205 J = 15.363 g X c X (27.2 - 100.3 C) c = 0.183 J/gC

Conclusion: 1. Use the given specific heat capacity values below to calculate the percent error of the experimental specific heat capacity that you determined in Part I of the lab.

Known specific heat values — Iron: 0.444 J/g°C; Zinc: 0.390 J/g°C; Copper: 0.385 J/g°C, Aluminum: 0.900 J/g°C experimental - actual value x 100 % actual value experimental - actual value | x 100 % actual value - experimental value = 0.39 actual value (known) = 0.39 Now plug in.