Answer:
hypoeutectoid
Explanation:
ferrite: pure form of iron
cementite: It is iron carbide with 93.3% iron and 6.67% carbon
hypoeutectoid: Eutectoid steel with carbon fraction less than 0.8%
hypereutectoid: Eutectoid Steel with carbon content more than 0.8%
For the mentioned iron-carbide alloy,
% of carbon in iron-carbide alloy= percentage of cementite × percentage of carbon in cementite
% of carbon in iron-carbide alloy= 0.09× 0.0667
= 0.6%
so the alloy is hypoeutectoid
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
Just took the test and it ISN'T 92.6 V, my guess is 80 V but I dont know because it only says if I got it wrong.