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3 years ago

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The specific heat of a certain type of metal is 0.128 j/(g·°c). what is the final temperature if 305 j of heat is added to 64.6

g of this metal initially at 20.0 °c?

1 answer:

Lesechka [4]3 years ago

4 0

When we add a certain amount of heat Q to a substance, the temperature of the substance increases by a $\Delta T$ given by
$Q=m C_s \Delta T$
where m is the mass of the substance and Cs is the specific heat capacity of the substance.

By re-arranging the formula, we find
$\Delta T = \frac{Q}{m C_s}= \frac{305.0 J}{(64.6 g)(0.128 J/gC)}=36.9 ^{\circ}C$

So, since the initial temperature of the metal is Ti=20 C, the final temperature is
$T_f = T_i + 36.9 ^{\circ} C=20.0^{\circ} + 36.9^{\circ} C=56.9^{\circ}C$

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On a hot summer day, 3.50 ✕ 106 J of heat transfer into a parked car takes place, increasing its temperature from 36.5°C to 44.4

anygoal [31]

Answer:

a) $\Delta s=443037.9747\ J.K^{-1}$

b) $\Delta s=31868131.8681\ J.K^{-1}$

Explanation:

a)

Given:

amount of heat transfer occurred, $dQ=3.5\times 10^6\ J$

initial temperature of car, $T_i=36.5+273=309.5\ K$

final temperature of car, $T_f=44.4+273=317.4\ K$

We know that the change in entropy is given by:

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)}$ (heat is transferred into the system of car)

$\Delta s=443037.9747\ J.K^{-1}$

b)

amount of heat transfer form the system of house, $dQ=5.8\times 10^8\ J$

initial temperature of house, $T_i=23.5+273=296.5\ K$

final temperature of house, $T_f=5.3+273=278.3\ K$

$\Delta s=\frac{dQ}{T_f-T_i}$

$\Delta s=\frac{5.8\times 10^8}{278.3-296.5}$

$\Delta s=31868131.8681\ J.K^{-1}$

6 0

3 years ago

WILL MARK BRAINLIEST!!!!

zloy xaker [14]

Answer:

Wave model

Explanation:

7 0

3 years ago

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Transverse, surface, and longitudinal waves are all __________ waves because they __________.

algol [13]

Answer:

a. mechanical; require a medium to travel through

Explanation:

Longitudinal, transverse and surface waves are types of mechanical waves. For example, within the longitudinal waves are the sound waves, which needs a medium to propagate like the air. This is why sound does not travel in a vacuum.

And an example of a transverse wave is the waves that form in the water when a rock is thrown (ripples), these waves need a medium (the water) to propagate.

On the other hand, electromagnetic waves such as light waves do not need a medium to propagate, this is why we can see the light of distant stars because their light travels through the vacuum until it reaches us.

So, the answer is:

Transverse, surface, and longitudinal waves are all mechanical waves because they require a medium to travel through .

5 0

3 years ago

Two loudspeakers emit sound waves along the x-axis. A listener in front of both speakers hears a maximum sound intensity when sp

grandymaker [24]

Answer:

frequency of the sound = f = 1,030.3 Hz

phase difference = Φ = 229.09°

Explanation:

Step 1: Given data:

Xini = 0.540m

Xfin = 0.870m

v = 340m/s

Step 2: frequency of the sound (f)

f = v / λ

λ = Xfin - Xini = 0.870 - 0.540 = 0.33

f = 340 / 0.33

f = 1,030.3 Hz

Step 3: phase difference

phase difference = Φ

Φ = (2π/λ)*(Xini - λ) = (2π/0.33)* (0.540-0.33) = 19.04*0.21 = 3.9984

Φ = 3.9984 rad * (360°/2π rad)

Φ = 229.09°

Hope this helps!

5 0

3 years ago

A train, traveling at a constant speed of 22.0 m/s, comes to an incline with a constant slope. While going up the incline, the t

Charra [1.4K]

Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

t is time

V = at + U

using equation v - u = at to get line equation for the graph of the motion of the train on the incline plane

$V_{x} = mt + V_{o}$ where m is the slope

Comparing equation (1) and (2)

$V = V_{x}$

a = m

$U = V_{o}$

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

a = - 1.40 m/s²

Sunstituting a = - 1.40 m/s² and u = 22 m/s

$V_{x} = -1.40t + 22$

$V_{x} = -1.40(7.30) + 22$

$V_{x} = -10.22 + 22$

$V_{x} = 11. 78 m/s$

The speed of the train at 7.30 s is 11.78 m/s.

The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.

Area of triangle + Area of rectangle

$[\frac{1}{2} * (22 - 11.78) * (7.30)] + [(11.78 - 0) * (7.30)]$

= 37.303 + 85.994

= 123. 297 m

≈ 123. 30 m

4 0

3 years ago

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