Describe the motion of an object that has an elephant acceleration of 0 mi./s squared

An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su

s2008m [1.1K]

Answer: A. 23.59 minutes.

B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

Latent heat of fusion is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

Specific heat capacity of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, given that:

mass of ice, m= 0.555 kg

temperature of ice, T= -16.6°C

rate of heat transfer, $q=820 J.min^{-1}$

specific heat of ice, $c_{i}= 2100 J.kg^{-1}.K^{-1}$

latent heat of fusion of ice, $L_{i}=334\times10^{3}J.kg^{-1}$

Asked:

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

We have the formula:

$Q=mc\Delta T$

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

$Q= 0.555\times2100\times16.6$

$Q= 19347.3 J$

Now, we require 19347.3 joules of heat to bring the ice to 0°C and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

$t=\frac{Q}{q}$

$=\frac{19347.3}{820}$

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

Now comes the concept of Latent heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

We have the equation: latent heat, $Q_{L}= mL_{i}$

$Q_{L}= 0.555\times334\times10^{3}= 185370 J$

Now the time required for supply of 185370 J:

$t=\frac{Q_{L}}{q}$

$t=\frac{185370}{820}$

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

8 0

3 years ago

Electrons in a particle beam each have a kinetic energy of 4.0 × 10−17 J. What is the magnitude of the electric field that will

denpristay [2]

Explanation:

Relation between work and change in kinetic energy is as follows.

$W_{net} = \Delta K$

Also, $\Delta K = K_{initial} - K_{final}$

= $(0 - 4.0 \times 10^{-17})$ J

= $-4.0 \times 10^{-17}$ J

Let us assume that electric force on the electron has a magnitude F. The electron moves at a distance of 0.3 m opposite to the direction of the force so that work done is as follows.

w = -Fd

$-4.0 \times 10^{-17} J = -F \times 0.3 m$

F = $1.33 \times 10^{-16}$

Therefore, relation between electric field and force is as follows.

E = $\frac{F}{q}$

= $\frac{1.33 \times 10^{-16}}{1.60 \times 10^{-19} C}$

= $0.831 \times 10^{3}$ C

Thus, we can conclude that magnitude of the electric field that will stop these electrons in a distance of 0.3 m is $0.831 \times 10^{3}$ C.

3 0

4 years ago

The frequency of the sound waves produced by a string _____.

Gnoma [55]

B.

increases as the tension of the string increases

7 0

3 years ago

Read 2 more answers

Suppose a two-level system is in a bath with temperature 247 K. The energy difference between the two states is 1.1 × 10-21 J. W

Alik [6]

Answer:

The probability of higher energy state is 0.4200.

Explanation:

Given that,

Temperature = 247 K

Energy difference between two states $E_{2}-E_{1}=1.1\times10^{-21}\ J$

We need to calculate the probability of higher energy state

Probability of $E_{1}= e^{-\beta E_{1}}$

Probability of $E_{2}= e^{-\beta E_{2}}$

The total probability is

$e^{-\beta E_{1}}+e^{-\beta E_{1}}=1$

Here, E₁ = lower energy state

E₂ = higher energy state

Put the value of E₁ in to the formula

$e^{-\beta(E_{2}-1.1\times10^{-21})}+e^{-\beta E_{1}}=1$

$e^{-\beta E_{2}}(e^{\beta 1.1\times10^{-21}}+1)=1$

$e^{\beta E_{2}}=\dfrac{1}{1+e^{\dfrac{1.1\times10^{-21}}{KT}}}$

Here, $\beta=\dfrac{1}{KT}$

Put the value into the formula

$e^{\beta E_{2}}=\dfrac{1}{1+e^{\dfrac{1.1\times10^{-21}}{1.380\times10^{-23}\times247}}}$

$e^{\beta E_{2}}=0.4200$

Hence, The probability of higher energy state is 0.4200.

4 0

3 years ago

While chatting with a friend you place your book bag on a nearby slide in the playground at school. The bag remains stationary.

alukav5142 [94]

Answer:

3. fs < μmg

4. fs = mg sinθ

Explanation:

For any object placed on a slide, there are 3 external forces acting on it:

Fg = m*g (always downward)
N (normal force, always perpendicular to the surface of the slide. going upward)
Fs (Friction Force, always opposite to the movement of the object, parallel to the slide)

As we have only one force with components along the normal and parallel to the slide directions (gravity force), it is advisable to find the components of this force, along these directions.

If θ is the angle of the slide above the horizontal, we have the following components of Fg:

Fgn = m*g*cosθ

Fgp = m*g*sin θ

We can apply Newton's 2nd Law to these perpendicular directions:

Fp = m*g*sin θ - Fs

Fn = N -m*g*cosθ = 0 (as the object has no movement in the direction perpendicular to the slide) (1)

Looking at the equation for the parallel direction, we have two forces, the component of Fg along the slide (which tries to accelerate the object towards the bottom of the slide), and the friction force.

While the object remains stationary, the equation for Newton's 2nd Law along this direction is as follows:

m*g*sin θ - fs =0 ⇒ fs = m*g*sinθ (4.)

This force can take any value (depending on the angle θ) to equilibrate the component of Fg along the slide, up to a limit value, which is given by the following expression:

fsmax = μN (2)

From (1), N= m*g*cos θ

Replacing in (2):

fsmax = μ*m*g*cos θ

While the bag remains at rest, we can say:

fs < μ*m*g*cosθ < μ*m*g (as in the limit cosθ =1)

So, the following is always true:

fs < μmg (3.)

6 0

3 years ago