Explanation:
increase disc radius
increase caliper piston area
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
Answer:
Explanation:
Detailed explanation given in the attached document.
Answer: Extension is Final length - initial length. This implies : 31.6-31 =0.6. Therefore the extension is 0.6
Explanation:
Answer:
d/dt[mCp(Ts-Ti)] = FCp(Ts-Ti) - FoCp(Ts-Ti) + uA(Ts-Ti)
Explanation:
Differential balance equation on the tank is given as;
<h3>Accumulation = energy of inlet steam - energy of outlet steam+ </h3><h3> heat transfer from the steam</h3><h3>where; </h3>
Accumulation = d/dt[mcp(Ts-Ti)]
Energy of inlet steam = FCp(Ts-Ti)
Energy of outlet steam = FoCp(Ts-Ti)
Heat transfer from the steam = uA(Ts-Ti)
Substituting into the formula, we have;
<h3>Accumulation = energy of inlet steam - energy of outlet steam+ </h3><h3> heat transfer from the steam</h3><h3 /><h3>d/dt[mCp(Ts-Ti)] = FCp(Ts-Ti) - FoCp(Ts-Ti) + uA(Ts-Ti)</h3><h3 /><h3 />