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)

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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.

(a) The specific heat capacity of the unknown sample is 1.82 J/g⁰C.

(b) From the value of the specific heat capacity, the material made of Beryllium.

Specific heat capacity of the unknown sample

The specific heat capacity of the unknown sample is calculated as follows;

Heat lost by the samples = Heat gained by the calorimeter and water

[tex]M_cC_c\Delta \theta_c + M_sC_s\Delta \theta_s = (M_wC_w + M_a_lC_a_l)\Delta \theta[/tex]

300(0.385)(30 - 20) + 70Cs(100 - 20) = (250 x 4.184  +  100 x 0.89)(20 - 10)

1155 + 5600Cs = 11350

5600Cs = 10,195

Cs = 1.82 J/g⁰C

Thus, from the value of the specific heat capacity, the material made of Beryllium.

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