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Oksi-84 [34.3K]
3 years ago
5

Five kg of water is contained in a piston-cylinder assembly, initially at 5 bar and 240°C. The water is slowly heated at constan

t pressure to a final state. If the heat transfer for the process is 2960 kJ, determine the temperature at the final state, in °C, and the work, in kJ. Kinetic and potential energy effects are negligible.
Engineering
1 answer:
Digiron [165]3 years ago
7 0

Answer:

The final temperature of water is 381.39  °C.

Explanation:

Given that

Mass of water = 5 kg

Heat transfer at constant pressure Q = 2960 KJ

Initial temperature = 240 °C

We know that heat transfer at constant pressure given as follows

Q=mC_p\Delta T

We know that for water

C_p=4.187\ \frac{KJ}{kg.K}

Lets take final temperature of water is T

So

Q=mC_p\Delta T

2960=5\times 4.187(T-240)

T=381.39  °C

So the final temperature of water is 381.39  °C.

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See the explanation for the answer;

Explanation:

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-------------------------------------------------------------------------------------Start of code

% Program to create a magic square and verify it

clc

M= magic(5)

% To find sum of elements of each row

r1= sum(M(1,:))      % Sum of row 1

r2= sum(M(2,:))      % Sum of row 2

r3= sum(M(3,:))      % Sum of row 3

r4= sum(M(4,:))      % Sum of row 4

r5= sum(M(5,:))      % Sum of row 5

% To find sum of each coloumn

c1= sum(M(:,1))      % Sum of coloumn 1

c2= sum(M(:,2))      % Sum of coloumn 2

c2= sum(M(:,3))      % Sum of coloumn 3

c2= sum(M(:,4))      % Sum of coloumn 4

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% To find sum of diagonal

d1= sum(diag(M))               % Sum of principal diagonal elements

d2= sum(diag(fliplr(M)))      

-------------------------------------------------------------------------------End of code

Following results are obtained when executed.

M =

   17    24     1     8    15

   23     5     7    14    16

    4     6    13    20    22

   10    12    19    21     3

   11    18    25     2     9

r1 =

   65

r2 =

   65

r3 =

   65

r4 =

   65

r5 =

   65

c1 =

   65

c2 =

   65

c2 =

   65

c2 =

   65

c2 =

   65

d1 =

   65

d2 =

   65

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