All categories

2 years ago

a heater 60w evaporates 6×10–3kg of boiling water in 60s.what is the specific latent heat of vaporisation of water in jkg–1

Physics

1 answer:

geniusboy [140]2 years ago

4 0

Your question doesn't look right, so lemme assume you meant this.

A heater marked 60w evaporate 6x 10^-³kg of boiling water in 60 seconds. What is the specific latent heat of vaporization of water in jkg?

so you can understand it better..

Latent heat of evaporation is the heat required required to change water to vapor at the same temperature.(100°C)

eg. water boils at 100°C and in presence of more heat it turns to vapor at that Same temperature.

This heat is know as latent heat of Vaporization.

it's give by H=mL

where L is the specific latent heat of vaporization.

You might be interested in

Vector A⃗ points in the negative y direction and has a magnitude of 5 km. Vector B⃗ has a magnitude of 15 km and points in the p

Alexxandr [17]

Answer:

magnitude of A − B = 15.81 km

Explanation:

Vector A points in the negative y-direction and has a magnitude of 5 km. Vector B points in the positive x-direction and has a magnitude of 15 km.

According to Cartesian coordinate system, the resultant will start either from tail of A and ends at head of B and vice-versa.

A(0,-5)

B(15,0)

A - B = (-15 i - 5 j )

Magnitude of the vector is given by

|A - B| = $\sqrt{(-15)^{2}+(-5)^{2}}$

|A - B| = $\sqrt{250}$

|A - B| = 15.81 km

7 0

3 years ago

A coaxial cable consists of a solid inner cylindrical conductor of radius 2 mm and an outer cylindrical shell of inner radius 3

4vir4ik [10]

Answer:

d) 1.2 mT

Explanation:

Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.

First of all, we observe that:

- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is

I = 15 A

- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).

Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

$B=\frac{\mu_0 I}{2\pi r}$

where

$\mu_0$ is the vacuum permeability

I = 15 A is the current in the conductor

r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field

Substituting, we find:

$B=\frac{(4\pi\cdot 10^{-7})(15)}{2\pi(0.0025)}=1.2\cdot 10^{-3}T = 1.2 mT$

8 0

2 years ago

Bodies A and B have equal mass. Body B is initially at rest. Body A collides with body B in a one-dimensional elastic collision.

jek_recluse [69]

According to the statement " Collision between two bodies in which the total kinetic energy of the two bodies after the collision is equal to their total kinetic energy before the collision."
The best answer is :
Option A " BODY A COMES TO REST BODY B STARTS MOVING WITH INITIAL VELOCITY OF BODY A "

4 0

2 years ago

Read 2 more answers

An electric heater containing two heating wires X and Y is connected to a power supply of electromotive force(emf) 9.0V and negl

mina [271]

Answer:

0.4 ohms.

Explanation:

From the circuit,

The voltage reading in the voltmeter = voltage drop across each of the parallel resistance.

1/R' = 1/R1+1/R2

R' = (R1×R2)/(R1+R2)

R' = (2.4×1.2)/(2.4+1.2)

R' = 2.88/3.6

R' = 0.8 ohms.

Hence the current flowing through the circuit is

I = V'/R'................ Equation 1

Where V' = voltmeter reading

I = 6/0.8

I = 7.5 A

This is the same current that flows through the variable resistor.

Voltage drop across the variable resistor = 9-6 = 3 V

Therefore, the resistance of the variable resistor = 3/7.5

Resistance = 0.4 ohms.

7 0

2 years ago

Radar uses radio waves of a wavelength of 2.2 m . The time interval for one radiation pulse is 100 times larger than the time of

liq [111]

Answer:

The shortest distance to an object that this radar can detect would be

111 m

Explanation:

The shortest distance is the minimum distance that would be detected by the radar, The time of oscillation can be obtained thus;

T = 1/f ..........1

but f= v/λ substituting f into equation 1 we have;

T = λ/v...............2

Where λ is the wavelength = 2.2 m

v is the velocity of light since the radar is an electromagnetic wave

= 3 x $10^{8}$ m/s

T = 2.2 m / 3 x $10^{8}$ m/s

T = 7.33 x $10^{-9}$ s

Calculating the time interval required by the pulse

Since the interval of the pulse (t) is 100 times greater than the period of oscillation, the time of the pulse is expressed as;

t = 100 x T

t = 100 x 7.33 x $10^{-9}$ s

t = 7.33 x $10^{-7}$ s

Therefor the time of the pulse is 7.33 x $10^{-7}$ s

Calculating for the shortest distance of the radar

The shortest distance of the radar can be obtained using the equation below;

λ_s = (t/2) x v ............3

Substituting into equation 3 we have

λ_s = (7.33 x $10^{-7}$ /2) x 3 x $10^{8}$ m/s

λ_s = 111 m

Therefore the shortest distance to an object that this radar can detect would be 111 m

8 0

3 years ago

a heater 60w evaporates 6×10–3kg of boiling water in 60s.what is the specific latent heat of vaporisation of water in jkg–1​

a heater 60w evaporates 6×10–3kg of boiling water in 60s.what is the specific latent heat of vaporisation of water in jkg–1