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3 years ago

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what is the minimum heat if A 500cm³ of water is to be heated from a room temperature (28°C) to 100°C in order to prepare hot cu

p of coffee

1 answer:

mr Goodwill [35]3 years ago

7 0

I need my points my guy

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PLEASE HELP ME :)

ddd [48]

A) 350 J

- The initial internal energy of the cup is

$U_i = 230 J$

- The final internal energy of the cup is

$U_f = 580 J$

According to the first law of thermodynamics:

$U_f - U_i = Q+W$

where

Q is the heat absorbed by the system

W is the work done on the system

The work done on the system in this case is 0, so we can rewrite the equation as

$U_f - U_i = Q$

And so we find the heat transferred

$Q=580 J - 230 J=350 J$

B) IN the cup

Explanation:

in this situation, we see that the internal energy of the cup increases. The internal energy of an object/substance is proportional to its temperature, so it is a measure of the average kinetic energy of the molecules of the object/substance. Therefore, in this case, the temperature (and the energy of the molecules of the substance) has increased: this means that heat has been transferred INTO the system from the environment (the heat came from the sun).

8 0

3 years ago

A person pulls a crate of mass M = 63 kg a distance 40.0 m along a horizontal floor by a constant force FP = 130 N, which acts a

GrogVix [38]

Answer:

Check attachment for solution and diagrams

Explanation:

Given that,

Mass of crate m=63kg

Distance travelled d=40m

Horizontal force Fx=130N

Angle the force applied on cord makes with horizontal is θ=23°.

The weight of the crate is given by

W=mg

W=63×9.81

W=618.03N

Horizontal force Fx=130N

Resolving the applied Force F to the horizontal will give

Fx=FCos θ

F=Fx/Cos θ

F=130/Cos23

F=141.2N

a. Check attachment for model diagram

b. Check attachment for free body diagram

c. Check attachment for pictorial representation

d. Work done by gravitational force.

We, know that the body did not move upward, then the distance d=0

Work done is given as

W=F×d

So, d=0

W=F×0

W=0J

So, no work is done by gravity

e. Normal force?

Using newton law of motion

ΣFy = may

Since the body is not moving upward, then ay=0m/s²

N+141.2Sin23-618.03=0

N=618.03-141.2Sin23

N=562.86N

f. Work done by normal force.

The body is not moving upward, then the distance is zero

d=0

Work done by normal=normal force × distance

Wn=562.86×0

Wn=0J

No work is done by the normal force

g. Frictional force?

Since the coefficient of kinetic friction is zero, then the surface is frictionless

So, no frictional force is acting on the body

Fictional force is given as

Fr=μk•N

Given that, μk=0

Fr=0×562.86

Fr=0N

d. Work done by frictional force?

Since the frictional force Is zero, then, no work is done by friction

W(friction ) = frictional force × d

Here, the body moved a distance of 40m

W(fr)=0×40

W(fr)=0J

No work is done by friction

I. Work done by exerted force

The horizontal component of the exerted force is 130N and the body traveled a distance of 40m

Then, work done is given as

Workdone=force ×distance

Work done=130×40

W=5200J

W=5.2KJ

h. Net workdone?

Since no work is lost by friction, then, the net work done is equal to the work done by the exerted force.

Went, = work done by force exerted - work done by friction

Wnet=5200-0

Wnet, =5200

Wnet=5.2KJ

7 0

3 years ago

Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity i

leonid [27]

Answer:

Explanation:

To find the amplitude of the sound, we must first determine the wavelength and the phase difference between the two speakers.

For the wavelength;

Recall that, the separation between two successive max. and min. intensity points are $\dfrac{\lambda}{2}$

Thus; for both speakers; the wavelength of the sound is:

$\dfrac{\lambda}{2} = (10+30) cm$

$\dfrac{\lambda}{2} = (40) cm$

λ = 80 cm

The relation between the path difference(Δx) and the phase difference(Δ∅) is:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

where;

Δx = 10 cm

λ = 80 cm

Δ∅ = π rad

∴

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{80 \ cm}(10 \ cm) + \Delta \phi_o$

$\pi \ rad = \dfrac{2 \pi}{8}+ \Delta \phi_o$

$\pi \ rad = \dfrac{ \pi}{4}+ \Delta \phi_o$

$\Delta \phi_o = \pi -\dfrac{ \pi}{4}$

$\Delta \phi_o = \dfrac{ 4\pi - \pi}{4}$

$\Delta \phi_o = \dfrac{ 3\pi}{4} \ rad$

Suppose both speakers are placed side-by-side, then the path difference between the two speakers is: Δx = 0 cm

Thus, we have:

$\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o$

$\Delta \phi = \dfrac{2 \pi}{\lambda}(0 \ cm ) + \dfrac{3 \pi}{4} \ rad$

$\Delta \phi = \dfrac{3 \pi}{4} \ rad$

∴

The amplitude of the sound wave if the two speakers are placed side-by-side is:

$A = 2a \ cos \bigg (\dfrac{\Delta \phi }{2} \bigg)$

$A = 2a \ cos \bigg (\dfrac{\dfrac{3 \pi}{4} }{2} \bigg)$

$A = 2a \ cos \bigg ({\dfrac{3 \pi}{8} } \bigg)$

A = 0.765a

7 0

3 years ago

The thermal transfer of energy by electromagnetic waves through the vacuum of space is called

Olenka [21]

Answer: Radiation

Explanation:

There are three ways in which the thermal transfer occurs:

1. By Conduction, when the transmission is by the <u>direct contact</u>.

2. By Convection, heat transfer <u>in fluids </u>(like water or the air, for example).

3. By Radiation, by the <u>electromagnetic waves</u> (they can travel through any medium and in <u>vacumm</u>)

So, in the outter space is vacuum, this means the energy cannot be transmitted by convection, nor conduction. It must be transmitted by electromagnetic waves that are able to travel with or without a medium, and this is called radiation.

4 0

3 years ago

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According to the theory of global warming, the average temperature of Earth’s air began rising rapidly when humans first started

ELEN [110]

I'd say the answer to this on is d.Facts as they are using the temperatures in the graph and the temperatures are not just estimates

3 0

3 years ago

Read 2 more answers