Air is compressed from 100 kPa, 300 K to 1000 kPa in a two-stage compressor with intercooling between stages. The intercooler pr

Answer:

final temperature = 26.5°

Explanation:

The water's initial volume is calculated as 1 x 1 x 1 = 1

The starting temperature of the water is 20° C

Water density = 1000 kg/

Volume of the copper block is determined as 0.46 x 0.46 x 0.46 = 0.097

The copper block's initial temperature is 100° C

Copper density = 8960 kg/

The water's final volume after accounting for the copper block is 1 - 0.097 = 0.903

Assumptions:

• the tank is adiabatic, meaning there’s no heat exchange through its walls
• the tank is completely filled, lacking space for any air to cool the water
• the total thermal energy in the tank is comprised of both the water’s heat energy and that of the copper block.

The mass of the remaining water in the tank can be calculated using density x volume = 1000 x 0.903 = 903 kg

The specific heat capacity for water, c = 4186 J/K-kg

Total heat content of the water Hw = mcT = 903 x 4186 x 20 = 75.59 Mega-joules

The copper's mass is calculated as density x volume = 8960 x 0.097 = 869.12 kg

Copper's specific heat capacity is 385 J/K-kg

The heat content of copper Hc = mcT = 869.12 x 385 x 100 = 33.46 Mega-joules

The overall heat in the system totals up to 75.59 + 33.46 = 109.05 Mega-joules

This heat will be evenly dispersed across the system

The heat energy of the water in the system is expressed as mcT

where T signifies the final temperature

= 903 x 4186 x T = 3779958T

For copper, the heat will be

mcT = 869.12 x 385 = 334611.2T

The combined heat from both will equal the total heat of the system, meaning

3779958T + 334611.2T = 109.05 x

4114569.2T = 109.05 x

Thus, the final temperature T = (109.05 x )/4114569.2 = 26.5°

Answer:

90 L.atm

Explanation:

According to the provided details:

First pathway:

A( 3 atm, 20 L) → C ( 1 atm, 20 L) → D (1 atm, 50 L)

Second pathway:

A(3 atm, 20 L) → B( 3 atm, 50 L) → D ( 1 atm, 50 L)

As the number of moles is 1.00 moles

To calculate wAB;

A → B signifies the transformation is happening at a steady pressure;

Thus,

wAB = pressure multiplied by the change in volume

wAB = P(V₂ - V₁)

wAB = 3 atm (50 L - 20 L)

wAB = 3 atm (30 L)

wAB = 90 L.atm

Answer:

The Python code is provided below with suitable comments

Explanation:

# convert list to set

male_names = set(['Oliver','Declan','Henry'])

# get name for removal and addition from user

remove_name = input("Enter remove name: ")

add_name = input("Enter add name: ")

# remove specified name from set

male_names.remove(remove_name)

# add new name to set

male_names.add(add_name)

# sort the resulting set

a = sorted(male_names)

# output the sorted set

print(a)

Answer:

The resolution for this question is below in the explanation section.

Explanation:

The right response to this question is A, which is 112002.

The correct code relating to this question is presented below

#include  <stdio.h>

#include  <string.h>

#include  <sys/types.h>

//#include "csapp.h"

void end(void)

{

printf("2");

}

int main()

{

if (fork() == 0)

atexit(end);

if (fork() == 0)

printf("0");

else

printf("1");

exit(0);

}

/* $end forkprob2 */

When this program is executed, the fork function will yield varying results.

However, it should be noted that running it multiple times will produce inconsistent values. The most common output the program will generate is A.

Attached is an image of the program's execution for further clarification.

It is not feasible to install the wires within the conduit. Explanation: The given dimensions show that the total area is 2.04 square inches while the wires occupy 0.93 square inches. The maximum allowable fill for the conduit is 40%. To determine if placement is possible, compute the conduit’s area at 40% which equates to 0.816 square inches, less than the required area of the wires.