100 g of iron shot is placed in a test tube and the test tube is placed in a bea

Question

100 g of iron shot is placed in a test tube and the test tube is placed in a beaker with 150 ml of water at room temperature. The beaker/test tube combnation is heated until the temperature of the iron metal reaches 100 degrees C. The heated iron metal is poured in to a calorimeter with 100 ml of water at room temperature and the calorimeter is closed. The recorded temperature of the calorimeter is 28 degrees C.

1. Calculate the specific heat of the iron metal.

2. Calculate the atomoc weight of the iron metal.

Explanation / Answer

Assumption # room temp is 25 C

1.Initial temp of iron shots = 100 C ; Final temp with caloriemeter combo = 28 C

water density assumed 1g/ml => 100 ml of water weighs 100gm

heat lost by iron = heat gained by water in caloriemter

=> (100)(C)(100-28) = (100)(4.18)(28-25)

=> C = 0.17417 J/g C (ans)

2. By Dulong Petit's Law

Molar mass * specific heat = 25 J/kg C

=> Molar mass of iron is 25 /0.17417 = 143.54 g/mol

which is very different from the theoritcal value. However, my ans is correct with the data provided/ Do recheck the data for missing info

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