Procedure: Step 1: Obtain metal sample in test tube. First add a test tube to th

Question

Procedure:

Step 1: Obtain metal sample in test tube. First add a test tube to the Lab. Select the test tube and add 100 gm of Iron Shot (Fe). Add a thermometer to the test tube so that you can record the initial temperature of the Fe.

Step 2: Prepare hot water bath and add test tube with metal. Obtain a 250 ml beaker. Add 150ml of water at room temperature to the beaker. Place test tube within beaker (select both test tube and beaker, and combine from the Arrange Menu®Combine). Now heat the combined beaker test tube arrangement by heating with a bunsen burner until the Fe shot reaches the boiling point of water 100 °C.

Step 3: Place heated metal in calorimeter with water at room temp. Add Calorimeter to Lab (select Equipment menu®Calorimeter) Add 100 ml of water at room temp 20 °C to calorimeter (select calorimeter and press water toolbar button 2x or use water dialog box). Remove test tube from beaker (by selecting beaker and using the Arrange Menu®Remove menu). Pour the heated Iron into the calorimeter and close the calorimeter (close calorimeter by selecting calorimeter and pressing the thermometer tool bar button. This will add both a thermometer and a cork top to calorimeter). Record the final temperature of the water.

Step 4: Repeat steps 1 -3 with other metals, recording final temperature and compare.

Observations:

Initial temperature of        metal:

Final temperature of        metal:

Initial temperature of water from calorimeter:

Final water temperature from calorimeter:

Estimate of specific heat for metal:

(use 1 cal/gm-°C for water)

Estimate the atomic weight of         metal:

Explanation / Answer

Mass of iron shot = 100 g

Initial temperature of iron shot = 100 oC

Final temperature of iron shot = 28 oC

Volume of water = 100 mL

Mass of water = 100 g

Initial temperature of water = 20 oC

Final temperature of water = 28 oC

q released = q absorbed

100 * C * (100 - 28) = 100 * 1 * (28 - 20)

C * 72 = 8

C = 0.111 Cal / g oC

C = 0.460 J / g oC

specific heat for metal = 0.460 J / g oC

Using Dulong-Petit law

C/N = 3R

0.460 / N = 3 * 8.314

N = 0.0184 moles / g

Atomic mass = 1 / 0.0184

= 54.34 g / mole

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