Question

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)

Answer 1

Answer:

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.