A household refrigerator that has a power input of 450 W and a COP of 1.5 is to cool 5 large watermelons, 10 kg each, to 8 C. If the watermelons are initially at 28 C, determine how long it will take for the refrigerator to cool them.


Answer 1


\Delta t = 5866.667\,s\,(97.778\,m)


The specific heat for watermelon above freezing point is 3.96\,(kJ)/(kg\cdot K). The heat liberated by the watermelon to cool down to 8°C is:

Q_(cooling) = (5)\cdot (10\,kg)\cdot (3.96\,(kJ)/(kg\cdot K) )\cdot (20\,K)

Q_(cooling) = 3960\,kJ

The heat absorbed by the household refrigerator is:

\dot Q_(L) = COP\cdot \dot W_(e)

\dot Q_(L) = 1.5\cdot (0.45\,kW)

\dot Q_(L) = 0.675\,kW

Time needed to cool the watermelons is:

\Delta t = (Q_(cooling))/(\dot Q_(L))

\Delta t = (3960\,kJ)/(0.675\,kW)

\Delta t = 5866.667\,s\,(97.778\,m)


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What does Enter key do? You cannot click Enter key to start a line if your current is blank?

This is spot to do today



See below


Enter-key also called the "Return key," it is the keyboard key that is pressed to signal the computer to input the line of data or the command that has just been typed.It Was the Return KeyThe Enter key was originally the "Return key" on a typewriter, which caused the carriage to return to the beginning of the next line on the paper. In a word processing or text editing application, pressing Enter ends a paragraph. A character code for return/end-of-line, which is different in Windows than it is in the Mac, Linux or Unix, is inserted into the text at that point.




Once there are two yellow lines having inner broken lines on the two sides of a center traffic lane, what this is trying to tell you is that you can use those lanes to start a left hand turn, or a U-turn from the both directions of traffic. However you cannot use it for passing. This is sometimes misunderstood by road users and drivers.

A multipurpose transformer has a secondary coil with several points at which a voltage can be extracted, giving outputs of 5.60, 12.0, and 480 V. (a) The input voltage is 220 V to a primary coil of 230 turns. What are the numbers of turns in the parts of the secondary used to produce the output voltages




A multipurpose transformer can act as step up as well as step down transformer according to the desired setting by a user.

When the voltage at the output is greater than the voltage at the input of the transformer then it acts as step-up transformer and vice-versa acting is a step down transformer.

Given that:

input (primary) voltage of the transformer, V_i=220~V

no. of turns in the primary coil, N_i=230

  • When the output voltage is 5.60 V:


(N_i)/(N_o) =(V_i)/(V_o)


N_o=5.85\approx 6 turns compensating the losses

  • When the output voltage is 12.0 V:


(N_i)/(N_o) =(V_i)/(V_o)


N_o=12.45\approx 13 turns compensating the losses

  • When the output voltage is 480 V:


(N_i)/(N_o) =(V_i)/(V_o)


N_o=501.8\approx 502 turns compensating the losses

One kind of SS-3xX steel alloy has a melting point of 1450°c. Its specific heat = 0.46 J/g°C, and its heat of fusion 270 J/g. For a 200 kg block of this steel, determine how much heat is required to (a) raise its temperature from 25°C to its melting point and (b) transform it from solid to liquid phase.



a)Q=131.1 MJ

b)Q=54 MJ


Given that

Mass ,m=200 kg

Specific heat Cp=0.46 J/g°C

Cp=0.46 KJ/kg°C

Heat of fusion = 270 J/g

Heat of fusion = 270 KJ/kg

Melting point temperature = 1450°C


Initial temperature = 25°C

Final temperature=1450°C

Heat required to rise temperature from 25°C to 1450°C.

Q= m CpΔT

Q=200 x 0.46 x (1450-25) KJ

Q=131,100 KJ

Q=131.1 MJ


Heat required to transform from solid phase to liquid phase

Q= Mass x heat of fusion

Q=200 x 27 KJ

Q=54,000 KJ

Q=54 MJ

A insulated vessel s has two compartments separated by a membreane. On one side is 1kg of steam at 400 degC and 200 bar. The other side is evacuated . The membrane ruptures, filling the entire volume. The finial pressure is 100bar. Determine the final temperature of the steam and the volume of the vessel.



See explaination


See attachment for the detailed step by step solution of the given problem.

The resultant force is directed along the positive x axis and has a magnitude of 1330 N. Determine the magnitude of F_A. Express your answer to three significant figures and include the appropriate units. Determine the direction theta of F_A. Express your answer using three significant figures.



the magnitude of F_A is 752 N

the direction theta of F_A is 57.9°


Given that,

Resultant force = 1330 N in x direction

∑Fx = R

from the diagram of the question which i uploaded along with this answer

FB = 800 N

FAsin∅ + FBcos30 = 1330 N

FAsin∅ = 1330 - (800 × cos30)

FA = 637.18 / sin∅

Now ∑Fx = 0

FAcos∅ - FBsin30 = 0

we substitute for FA

(637.18 / sin∅)cos∅ = 800 × sin30

637.18 / 800 × sin30 = sin∅/cos∅

and we know that { sin∅/cos∅ = tan∅)

so tan∅ = 1.59295

∅ = 57.88° ≈ 57.9°


FA = 637.18 / sin∅

we substitute ∅

so FA = 637.18 / sin57.88

FA = 752 N

8.2.1: Function pass by reference: Transforming coordinates. Define a function CoordTransform() that transforms the function's first two input parameters xVal and yVal into two output parameters xValNew and yValNew. The function returns void. The transformation is new



The output will be (3, 4) becomes (8, 10)


#include <stdio.h>

//If you send a pointer to a int, you are allowing the contents of that int to change.

void CoordTransform(int xVal,int yVal,int* xNew,int* yNew){

*xNew = (xVal+1)*2;

*yNew = (yVal+1)*2;


int main(void) {

int xValNew = 0;

int yValNew = 0;

CoordTransform(3, 4, &xValNew, &yValNew);

printf("(3, 4) becomes (%d, %d)\n", xValNew, yValNew);

return 0;