Answer:
The initial temperature is 649 K (376 °C).
The final pressure is 0.965 MPa
Explanation:
From the ideal gas equation
PV = nRT
P is the initial pressure of water = 2 MPa = 2×10^6 Pa
V is intial volume = 150 L = 150/1000 = 0.15 m^3
n is the number of moles of water in the container = mass/MW = 1000 g/18 g/mol = 55.6 mol
R is gas constant = 8.314 m^3.Pa/mol.K
T (initial temperature) = PV/nR = (2×10^6 × 0.15)/(55.6 × 8.314) = 649 K = 649 - 273 = 376 °C
From pressure law,
P1/T1 = P2/T2
P2 (final pressure) = P1T2/T1
T2 (final temperature) = 40 °C = 40 + 273 = 313 K
P1 (initial pressure) = 2 MPa
T1 (initial temperature) = 649 K
P2 = 2 × 313/649 = 0.965 MPa
Answer:
<em>Written in Python</em>
def SumN(n):
total = 0
for i in range(1,n+1):
total = total + i
print("Total: ",total)
def SumNCubes(n):
total = 0
for i in range(1,n+1):
total = total + i**3
print("Cube: ",total)
n = int(input("User Input: "))
if n > 0:
SumN(n)
SumNCubes(n)
Explanation:
The SumN function is defined here
def SumN(n):
This line initializes the total to 0
total = 0
The following iteration compute the required sum
<em> for i in range(1,n+1):
</em>
<em> total = total + i
</em>
This line outputs the calculated sum
print("Total: ",total)
The SumNCubes function is defined here
def SumNCubes(n):
This line initializes the total to 0
total = 0
The following iteration compute the required sum of cubes
<em> for i in range(1,n+1):
</em>
<em> total = total + i**3
</em>
This line outputs the calculated sum of cubes
print("Cube: ",total)
The main starts here; The first line prompts user for input
n = int(input("User Input: "))
The next line checks if input is greater than 0; If yes, the two defined functions are called
if n > 0:
SumN(n)
SumNCubes(n)
Answer:
Code fixed below using Java
Explanation:
<u>Error.java
</u>
import java.util.Random;
public class Error {
public static void main(String[] args) {
final int MAXCHEESE = 10;
String[] names = new String[MAXCHEESE];
double[] prices = new double[MAXCHEESE];
double[] amounts = new double[MAXCHEESE];
// Three Special Cheeses
names[0] = "Humboldt Fog";
prices[0] = 25.00;
names[1] = "Red Hawk";
prices[1] = 40.50;
names[2] = "Teleme";
prices[2] = 17.25;
System.out.println("We sell " + MAXCHEESE + " kind of Cheese:");
System.out.println(names[0] + ": $" + prices[0] + " per pound");
System.out.println(names[1] + ": $" + prices[1] + " per pound");
System.out.println(names[2] + ": $" + prices[2] + " per pound");
Random ranGen = new Random(100);
// error at initialising i
// i should be from 0 to MAXCHEESE value
for (int i = 0; i < MAXCHEESE; i++) {
names[i] = "Cheese Type " + (char) ('A' + i);
prices[i] = ranGen.nextInt(1000) / 100.0;
amounts[i] = 0;
System.out.println(names[i] + ": $" + prices[i] + " per pound");
}
}
}
Answer:
Se obtendrán 116.66 litros de jugo concentrado, y el agua evaporada será por un total de 883.33 litros.
Explanation:
Dado que para conseguir jugo de naranja concentrada, se parte de un extracto con 7% en peso de sólidos el cual se mete a un evaporador al vacío, y en el evaporador se elimina el agua necesaria para que el jugo salga con una concentración del 60% de peso de sólidos, si se introducen al proceso 1000 kg/h de jugo diluido, para calcular la cantidad de agua evaporada y de jugo concentrado saliente se debe realizar el siguiente cálculo;
1000 x 0.07 = 70
60 = 70
100 = X
100 x 70 / 60 = X
7000 / 60 = X
116.66 = X
Por lo tanto, se obtendrán 116.66 litros de jugo concentrado, y el agua evaporada será por un total de 883.33 litros.
