An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10.0°C. Two metallic blocks are placed into the water. One is a 300.0-g piece of copper at 30.0°C. The other has a mass of 70.0 g and is originally at a temperature of 100°C. The entire system stabilizes at a final temperature of 20.0°C. (a) Determine the specific heat of the unknown sample? (b) Using the data in the table above, can you make a positive identification of the unknown material? (explain)

1822.14 J/kg C

Explanation:

Mass of aluminum = 0.100 kg

Mass of Water = 0.250 kg

Mass of copper = 0.300 kg

Mass of unknown object = 0.70 kg

Initial temperature of Aluminum = 10 C

Initial temperature of water = 10 C

Initial temperature of copper = 30 C

Initial temperature of = 100 C

Final temperature of all substances = 20 C

Change in temperature of each of them Aluminum, water, copper and unknown material respectively = 10 C, 10 C, - 10 C, - 80 C respectively

Specific heat of each of them Aluminum, water, copper and unknown material respectively = 900,4186,387 and c respectively.

Heat gained by aluminum and water = (0.1) (900) (10) + (0.250) (4186) (10) = 11365 j

heat lost by copper and unknown material = (0.3) (387) (-10) = - 1161

heat lost by unknown material = (0.070) (c) (-80) = - 5.6 c

Now solve for c using the calorimetry principle.

Heat lost + Heat gained = 0

⇒ - 5.6 c = - 11365 - (-1161) = - 10204 j

Specific heat of unknown material = - 10204 / - 5.6

= 1822.14 j/kg C

b) Material having a specific heat of 1800 j / kg C is

Polyurethane elastomer.