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adelina 88 [10]
3 years ago
12

How much heat is required to convert 0.3 kilogram of ice at 0°C to water at the same temperature

Physics
2 answers:
Grace [21]3 years ago
6 0

Answer:

100,200J of heat is required to convert 0.3kg of ice of 0°C to water at same temperature.

Explanation:

Heat = mass * lf

Latent heat of fusion (lf) of water is 334J/g

Heat = 300g * 334 J/g

Heat = 100,200J of heat

irina1246 [14]3 years ago
6 0

Answer:

100800 J

Explanation:

Specific Latent heat of fusion of water: This is the quantity of heat required to change the unit mass of water from ice to liquid without change in temperature.

The expression for Specific latent heat of fusion is given as

Q = Cm...................... Equation 1

Where Q = Amount of heat, C = Specific latent heat of fusion of ice, m = mass of ice.

Given: m = 0.3 kg,

Constant: C = 336000 J/kg

substitute into equation 1

Q = 336000(0.3)

Q = 100800 J.

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A photoelectric effect experiment finds a stopping potential of 1.93 V when light of wavelength 200 nm is used to illuminate the
GenaCL600 [577]

a) Zinc (work function: 4.3 eV)

The equation for the photoelectric effect is:

E=\phi + K (1)

where

E=\frac{hc}{\lambda} is the energy of the incident photon, with

h = Planck constant

c = speed of light

\lambda = wavelength

\phi = work function of the metal

K = maximum kinetic energy of the photoelectrons emitted

The stopping potential (V) is the potential needed to stop the photoelectrons with maximum kinetic energy: so, the corresponding electric potential energy must be equal to the maximum kinetic energy,

eV=K

So we can rewrite (1) as

E=\phi + eV

where we have:

\lambda=200 nm = 2\cdot 10^{-7} m

V = 1.93 V

e is the electron charge

First of all, let's find the energy of the incident photon:

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{2\cdot 10^{-7}m}=9.95\cdot 10^{-19} J

Converting into electronvolts,

E=\frac{9.95\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=6.22 eV

And now we can solve eq.(1) to find the work function of the metal:

\phi = E-eV=6.22 eV-1.93 eV=4.29 eV

so, the metal is most likely zinc, which has a work function of 4.3 eV.

b) The stopping potential is still 1.93 V

Explanation:

The intensity of the incident light is proportional to the number of photons hitting the surface of the metal. However, the energy of the photons depends only on their frequency, so it does not depend on the intensity of the light. This means that the term E in eq.(1) does not change.

Moreover, the work function of the metal is also constant, since it depends only on the properties of the material: so \phi is also constant in the equation. As a result, the term (eV) must also be constant, and therefore V, the stopping potential, is constant as well.

6 0
3 years ago
An inductance L, resistance R, and ideal battery of emf are wired in series. A switch in the circuit is closed at time t = 0, at
Kay [80]

Explanation:

After some time t the current does not passing through the circuit

=>so the back emf is zero

=>here the inductor opposes decay of the circuit

- Ldi/dt = Ri

di/dt = - R/Li

di/i = - R/Ldt

now we applying the integration on both sides

log i=-R/Lt+C

here t=0=>i=io

Log io=C

=>Log i=-R/L*t + Log io

logi-Log io=-R/L*t

Log[i/io]=-R/L*t

i/io=e^-Rt/L

i=ioe^-Rt/L

the option D is correct

3 0
3 years ago
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( WILL GIVE BRAINLIEST TO WHOEVER ANSWERS FIRST AND SHOWS WORK) A car weighs 15300 N. What is its mass?
elena-s [515]
The answer I received was 1561 kg

5 0
3 years ago
What caused the blackout in Canada?
Maru [420]

Answer:

Hello, the tripping of a 230-kilovolt transmission line.

Explanation:

the tripping of a 230-kilovolt transmission line near Ontario, Canada, at 5:16 p.m., which caused several other heavily loaded lines also to fail. Hopefully this helps you find what your looking for!.

7 0
3 years ago
You drive a car 1600 ft to the east, then 2500 ft to the north. The trip took 2.5 minutes. What was the magnitude of your averag
elena-14-01-66 [18.8K]

Answer:

Velocity=6.03m/s

Explanation:

Given data

Time t=2.5 minutes=150 seconds

Distance A=1600 ft=487.68 m........east

Distance B=2500 ft=762m ........north

To find

Average velocity

Solution

First we need to find the resultant distance magnitude.To find that we apply Pythagorean theorem to find hypotenuse

So

A^{2}+B^{2}=C^{2}\\  C=\sqrt{A^{2}+B^{2}}\\ C=\sqrt{(487.68m)^{2}+(762m)^{2}}\\ C=904.7m

Velocity=\frac{Distance}{Time}\\Velocity=\frac{904.7m}{150s}\\Velocity=6.03m/s

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