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3 years ago

A student attempted to measure the specific latent heat of vaporisation of water.

Physics

1 answer:

tensa zangetsu [6.8K]3 years ago

6 0

Answer:

The latent heat of vaporization of water is 2.4 kJ/g

Explanation:

The given readings are;

The first (mass) balance reading (of the water) in grams, m₁ = 581 g

The second (mass) balance reading (of the water) in grams, m₂ = 526 g

The first joulemeter reading in kilojoules (kJ), Q₁ = 195 kJ

The second joulemeter reading in kilojoules (kJ), Q₂ = 327 kJ

The latent heat of vaporization = The heat required to evaporate a given mass water at constant temperature

Based on the measurements, we have;

The latent heat of vaporization = ΔQ/Δm

∴ The latent heat of vaporization of water = (327 kJ - 195 kJ)/(581 g - 526 g) = 2.4 kJ/g

The latent heat of vaporization of water = 2.4 kJ/g

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Two stationary point charges of 3.00 nC and 2.00 nC are separated by a distance of 50.0 cm. An electron is released from rest at

andrew-mc [135]

Answer:

1. the electric potential energy of the electron when it is at the midpoint is - 2.9 x $10^{-17}$ J

2. the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge is - 5.04 x $10^{-17}$ J

Explanation:

given information:

$q_{1}$ = 3 nC = 3 x $10^{-9}$ C

$q_{2}$ = 2 nC = 2 x $10^{-9}$ C

r = 50 cm = 0.5 m

the electric potential energy of the electron when it is at the midpoint

potential energy of the charge, F

F = k $\frac{q_{e}q}{r}$

where

k = constant (8.99 x $10^{9} Nm^{2} /C^{2}$ )

electron charge, $q_{e}$ = - 1.6 x $10^{-19}$ C

since it is measured at the midpoint,

r = $\frac{0.5}{2}$

= 0.25 m

thus,

F = $F_{1}+ F_{2}$

= k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $\frac{kq_{e} }{r}$ ( $q_{1} +q_{2}$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ +2 x $10^{-9}$ )/0.25

= - 2.9 x $10^{-17}$ J

the electric potential energy of the electron when it is 10.0 cm from the 3.00 nC charge

$r_{1}$ = 10 cm = 0.1 m

$r_{2}$ = 0.5 - 0.1 = 0.4 m

F = k $\frac{q_{e} q_{1} }{r}$ + k $\frac{q_{e} q_{2} }{r}$

= $kq_{e}$ ( $\frac{q_{1} }{r_{1} }$ + $\frac{q_{2} }{r_{2} }$ )

= (8.99 x $10^{9}$ )( - 1.6 x $10^{-19}$ )(3 x $10^{-9}$ /0.1+2 x $10^{-9}$ /0.4)

= - 5.04 x $10^{-17}$ J

3 0

3 years ago

A vector is 14.4 m long and

MaRussiya [10]

Answer:

Explanation:

The x-component is found in the magnitude of the vector times the cosine of the angle.

$A_x=14.4cos133$ and, to 3 sig dig,

$A_x=-9.82m$

3 0

3 years ago

An electron in a mercury atom drops

aksik [14]

Since the electron dropped from an energy level i to the ground state by emitting a single photon, this photon has an energy of 1.41 × 10⁻¹⁸ Joules.

<h3>How to calculate the photon energy?</h3>

In order to determine the photon energy of an electron, you should apply Planck-Einstein's equation.

Mathematically, the Planck-Einstein equation can be calculated by using this formula:

E = hf

<u>Where:</u>

h is Planck constant.

f is photon frequency.

In this scenario, this photon has an energy of 1.41 × 10⁻¹⁸ Joules because the electron dropped from an energy level i to the ground state by emitting a single photon.

Read more on photons here: brainly.com/question/9655595

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4 0

2 years ago

(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .