In this experiment a hot metal block at 100 o C is added to cold water in a calo

Question

In this experiment a hot metal block at 100oC is added to cold water in a calorimeter. The heat lost by the metal block equals the heat gained by the water and the two end up at the same temperature.

In one experiment, the mass of the metal block is 79.9 grams, the volume of water used is 73.8 mL, the temperature of the cold water extrapolated to the time of mixing is 17.7oC, and the temperature of the metal block and water extrapolated back to the time of mixing is 28.7oC.

The specific heat of water is 4.184 J K-1g-1

How much heat is gained by the cold water?

Heat gained by cold water = J

What is the specific heat of the metal?

Specific heat of the metal = J K-1g-1

According to the Law of Dulong and Petit, what is the molar mass of the metal?

Molar mass =

Explanation / Answer

Mass of Water = 73.8 g

Sw = 4.184 J/ K g

Temperature change = Tw = 28.7 - 17.7 = 11 K

Heat Gained by Cold Water = 73.8 * 4.184 * 11 = 3396.57 J

The above heat gained by cold water is = heat lost by metal

Temperatue change of metal = Tm = 100 - 28.7 = 71.3 g

Mass of metal = 79.9 g

3396.57 = 79.9 * Sm * 71.3

which gives

Specific Heat of Metal = Sm = 0.596 J/ K g

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