Answer:
Q = -68.859 kJ
Explanation:
given details
mass 
initial pressure P_1 = 104 kPa
Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K
final pressure P_2 = 1068 kPa
Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K
we know that
molecular mass of 
R = 8.314/44 = 0.189 kJ/kg K
c_v = 0.657 kJ/kgK
from ideal gas equation
PV =mRT






WORK DONE

w = 586*(0.1033 -0.514)
W =256.76 kJ
INTERNAL ENERGY IS



HEAT TRANSFER

= 187.902 +(-256.46)
Q = -68.859 kJ
Answer:
(d) Spheroidizing
Explanation:
Spheroidizing
This is the heat treatment process for steel which having carbon percentage more than 0.8 %.As we know that a hard and brittle material is having carbon percentage more than 0.8 %.That is why this process is suitable for the hard materials.
In this process a hard and brittle materials convert into soft and ductile after this it improve the machine ability as well as improve the tool life.
In this process grain become spheroidal and these grains are ductile.
Answer:
Below is the desired C++ program for the problem. Do feel free to edit it according to your preference
Explanation:
#include <iostream>
#include <vector>
using namespace std;
void ExactChange(int userTotal, vector<int> &coinVals) {
coinVals.reserve(5);
coinVals[0] = userTotal / 100;
userTotal %= 100;
coinVals[1] = userTotal / 25;
userTotal %= 25;
coinVals[2] = userTotal / 10;
userTotal %= 10;
coinVals[3] = userTotal / 5;
userTotal %= 5;
coinVals[4] = userTotal;
}
int main() {
vector<int> coins;
int value;
cin >> value;
if (value <= 0) {
cout << "no change" << endl;
} else {
ExactChange(value, coins);
if (coins[0] != 0) cout << coins[0] << " " << (coins[0] == 1 ? "dollar" : "dollars") << endl;
if (coins[1] != 0) cout << coins[1] << " " << (coins[1] == 1 ? "quarter" : "quarters") << endl;
if (coins[2] != 0) cout << coins[2] << " " << (coins[2] == 1 ? "dime" : "dimes") << endl;
if (coins[3] != 0) cout << coins[3] << " " << (coins[3] == 1 ? "nickel" : "nickels") << endl;
if (coins[4] != 0) cout << coins[4] << " " << (coins[4] == 1 ? "penny" : "pennies") << endl;
}
return 0;
}