# Water is boiled at 1 atm pressure in a coffee maker equipped with an immersion-type electric heating element. The coffee maker initially contains 1 kg of water. Once boiling started, it is observed that half of the water in the coffee maker evaporated in 12 min. If the heat loss from the coffee maker is negligible, what is the power rating of the heating element?

1.57 KW

Explanation:

given data:

P= 1 atm

T= 12 min

power rating=??

solution:

latent heat of vaporization (L) of water at 1 atm = 2257.5 KJ/Kg

half of the water is evaporated in 12 min

so power rating is,

=P×L/2.T

=1×2257.5 /2×12×60

=1.57 KW

## Related Questions

Extra Credit: The Linc (parking lot and stadium)In celebration of the upcoming Super Bowl, for a maximum 10 points of extra credit, you may try to reproduce the ASCII Art shown below of Lincoln Stadium, home of the Philadelphia Eagles. You should still include a class constant for the SIZE; in Dr. Yates' implementation, the SIZE value that produces the picture below is 4, and works for any size >= 2. You must include loops and nested loops to make this work correctly; you CANNOT simply include a separate println statement for each line of the drawing. You will get the full extra credit points only if you duplicate the drawing EXACTLY. (Note: this is a fairly tricky figure to do right.) The parking lot alone is worth a maximum of 2 points.

Explanation:

// Below is the code to draw parking lot only , i have also put the o/p of code.

import java.util.*;

import java.lang.*;

import java.io.*;

public class Linc

{

static int size=4;

static int numBoxes = 1; // one row.

static int height =(int) Math.pow(size, 2); //Just how many | will it have.

static int width = 2; // two boxes in a row

public static void main (String[] args) throws java.lang.Exception

{

top();

printHeight();

}

public static void top(){

for(int i = 0; i

An excited electron in an Na atom emits radiation at a wavelength 589 nm and returns to the ground state. If the mean time for the transition is about 20 ns, calculate the inherent width in the emission line. What is the length of the photon emitted

Answer:   Inherent width in the emission line: 9.20 × 10⁻¹⁵ m or 9.20 fm

length of the photon emitted: 6.0 m

Explanation:

The emitted wavelength is 589 nm and the transition time is ∆t = 20 ns.

Recall the Heisenberg's uncertainty principle:-

∆t∆E ≈ h ( Planck's Constant)

The transition time ∆t corresponds to the energy that is ∆E

.

The corresponding uncertainty in the emitted frequency ∆v is:

∆v= ∆E/h = (5.273*10^-27 J)/(6.626*10^ J.s)=  7.958 × 10^6 s^-1

To find the corresponding spread in wavelength and hence the line width ∆λ, we can differentiate

λ = c/v

dλ/dv = -c/v² = -λ²/c

Therefore,

∆λ = (λ²/c)*(∆v) = {(589*10⁻⁹ m)²/(3.0*10⁸ m/s)} * (7.958*10⁶ s⁻¹)

=  9.20 × 10⁻¹⁵ m or 9.20 fm

The length of the photon (l) is

l = (light velocity) × (emission duration)

= (3.0 × 10⁸  m/s)(20 × 10⁻⁹ s) = 6.0 m

A pin-supported structure has unrestrained rotations at the support locations.a) True b) False

a)True

Explanation:

Yes it is true a pin support can not resist the rotation motion . It can resist only lateral or we can say that only linear motion of structure and can not resit angular moment of motion about hinge or pin joint.On the other hand a fixed support can resist linear as well rotation motion of structure.

Atmospheric air at 25 °C and 8 m/s flows over both surfaces of an isothermal (179C) flat plate that is 2.75m long. Determine the heat transfer rate per unit from the plate for 3 width different values of the critical Reynolds number: 100,000; 500,000; and 1,000,000

Re=100,000⇒Q=275.25

Re=500,000⇒Q=1,757.77

Re=1,000,000⇒Q=3060.36

Explanation:

Given:

For air      =25°C  ,V=8 m/s

For surface =179°C

L=2.75 m    ,b=3 m

We know that for flat plate

⇒Laminar flow

30\times10^5" alt="Re>30\times10^5" align="absmiddle" class="latex-formula">⇒Turbulent flow

Take Re=100,000:

So this is case of laminar flow

From standard air property table at 25°C

Pr= is 0.71  ,K=26.24

So

Nu=187.32   ()

187.32=

⇒h=1.78

heat transfer rate =h

=275.25

Take Re=500,000:

So this is case of turbulent flow

Nu=1196.18  ⇒h=11.14

heat transfer rate =h

=11.14(179-25)

= 1,757.77

Take Re=1,000,000:

So this is case of turbulent flow

Nu=2082.6  ⇒h=19.87

heat transfer rate =h

=19.87(179-25)

= 3060.36