A student carried out an experiment similar to this one. He found the temperatur

Question

A student carried out an experiment similar to this one. He found the temperature of the boiling water bath to be 100.7°C. Iron pieces weighing 51.70g were placed in a test tube and heated in the boiling water bath for 10 minutes. 50.0ml of water were placed in the calorimeter and the temperature was measured every 30 seconds for 4 minutes. The hot iron was dumped into the calorimeter and the readings were continued unatia a constant downward trend had been established. When the student extrapolated the temperatures to the mixing time he found that the water started at 22.2°C and ended up at 29.4°C.

Calculate the specific heat lost by the iron. Please show calculations.

Explanation / Answer

Solution:

According to the principle of the method of mixtures (heat transfer) , The heat lost by the hot body = heat gained by the cold body.

Here water is at the boiling temperature into which the iron pieces are dropped.Then the final equilibrium temperature  remained at 29.4 C.

Initial temperature of iron pieces =t iron = 100.7 C

Initial temperature of water = tw =22.2C

Mass of iron pieces = m iron =51.7 g = 0.0517kg

Mass of water = 5 ml = 5 e-3 kg

Final equilibrium temperature = t f = 29.4 C

Specific heat of86 J/kg C

Heat lost by iron = Heat gained by water

51.7 * Ciron (t iron - tf ) = 50 * 4186 (29.4 - 22.2)

=> SPecific heat of iron = 50 * 4186 (29.4 - 22.2) / (51.7 * (100.7 - 29.4)

=> C iron = 408.8 J/kg C

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