Answer:
Inspectors use inductive reasoning on the job.
Explanation:
I just took the test.
Answer:
a. Heat Capacity = 1.756J/mol-K
b. Heat Capacity = 24.942J/mol-k
Explanation:
Given
Constant volume Cv = 0.81J/mol-k
T1 = 34K
Td = Debye temperature = 306 K. Estimate the heat capacity (in J/mol-K) a. 44 K
First, The value of the temperature-independent constant.
Using Cv = AT³
Make A the subject of formula
A = Cv/T³
Substitute each values
A = 0.81/34³
A = 0.000020608589456543
A = 2.061 * 10^-5J/mol-k
The heat capacity changes with the temperature; below is the relationship between heat capacity and the temperature
Cv = AT³
So, The heat capacity when T = 44k is then calculated as
Cv = 2.061 * 10^-5 * 44³
Cv = 1.755522084266232
Cv = 1.756J/mol-K
(b) at 477 K.
Because the temperature is larger than the Debye temperature, the specific heat is calculated using as:
Cv = 3R
Where R = universal gas constant
R = 8.314J/mol-k
Cv = 3 * 8.314
Cv = 24.942J/mol-k
The impact behavior of plastic materials is strongly dependent upon the temperature. At high temperatures, materials are more ductile and have high impact toughness. At low temperatures, some plastics that would be ductile at room temperature become brittle.