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Akimi4 [234]
3 years ago
9

If a 20.0 g object at a temperature of 35.0∘C has a specific heat of 2.89Jg∘C, and it releases 450. J into the atmosphere, what

will be the final temperature of the object? Report your answer with the correct number of significant figures.
Physics
2 answers:
nataly862011 [7]3 years ago
8 0

Answer:

The final temperature of the object will be 42.785 °C

Explanation:

When the heat added or removed from a substance causes a change in temperature in it, this heat is called sensible heat.

In other words, sensible heat is the amount of heat that a body absorbs or releases without any changes in its physical state (phase change), so that the temperature varies.

The equation for calculating the heat exchanges in this case is:

Q = c * m * ΔT

where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature.

In this case:

  • Q= 450 J
  • c= 2.89 \frac{J}{g*C}
  • m= 20 g
  • ΔT= Tfinal - Tinitial= Tfinal - 35 °C

Replacing:

450 J= 2.89 \frac{J}{g*C} *20 g* (Tfinal - 35°C)

Solving for Tfinal:

\frac{450 J}{2.89\frac{J}{g*C}*20g} =Tfinal -35C

7.785 °C=Tfinal - 35°C

7.785 °C + 35°C= Tfinal

42.785 °C=Tfinal

<u><em>The final temperature of the object will be 42.785 °C</em></u>

lawyer [7]3 years ago
4 0

Answer: 27.2 degrees C

Explanation:

The final temperature can be determined by rearranging the specific heat capacity equation for ΔT and substituting the known values of mass, specific heat, and the heat released (which is represented by a negative number).

qcm=ΔT

−450 J / (2.89 Jg∘C)(20.0 g) = ΔT

-7.79∘C=ΔT

The negative sign of ΔT makes sense, as we know that energy is released in this exothermic process:

(Tinitial+ΔT=Tfinal)

35.0∘C+(–7.79∘C) = 27.2∘C

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9. A 5.0 kg block on an inclined plane is acted upon by a horizontal force of 100 N shown in the figure below. The coefficient o
Helga [31]

Answer:

A: The acceleration is 7.7 m/s up the inclined plane.

B: It will take the block 0.36 seconds to move 0.5 meters up along the inclined plane

Explanation:

Let us work with variables and set

m=5kg\\\\F_H=100N\\\\\mu=0.3\\\\\theta=37^o.

As shown in the attached free body diagram, we choose our coordinates such that the x-axis is parallel to the inclined plane and the y-axis is perpendicular. We do this because it greatly simplifies our calculations.

Part A:

From the free body diagram we see that the total force along the x-axis is:

F_{tot}=mg*sin(\theta)+F_s-F_Hcos(\theta).

Now the force of friction is F_s=\mu*N, where N is the normal force and from the diagram it is F_y=mg*cos(\theta).

Thus F_s=\mu*N=\mu*mg*cos(\theta).

Therefore,

F_{tot}=mg*sin(\theta)+\mu*mg*cos(\theta)-F_Hcos(\theta)\\\\=mg(sin(\theta)+\mu*cos(\theta))-F_Hcos(\theta).

Substituting the value for F_H,m,\mu, and \:\theta we get:

F_{tot}= -38.63N.

Now acceleration is simply

a=\frac{F_H}{m} =\frac{-38.63N}{5kg} =-7.7m/s.

The negative sign indicates that the acceleration is directed up the incline.

Part B:

d=\frac{1}{2} at^2

Which can be rearranged to solve for t:

t=\sqrt{\frac{2d}{a} }

Substitute the value of d=0.50m and a=7.7m/s and we get:

t=0.36s.

which is our answer.

Notice that in using the formula to calculate time we used the positive value of a, because for this formula absolute value is needed.

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3 years ago
URGENTTT PLEASE HELPPPP. You put m1 = 1 kg of ice cooled to -20°C into mass m2 = 1 kg of water at 2°C. Both are in a thermally i
STatiana [176]

Answer:

Explanation:

heat lost by water will be used to increase the temperature of  ice

heat gained by ice

= mass x specific heat  x rise in temperature

1 x 2090 x t

heat lost by water in cooling to 0° C

= mcΔt  where m is mass of water , s is specific heat of water and Δt is fall in temperature .

= 1 x 2 x 4186  

8372

heat lost = heat gained

1 x 2090 x t  = 8372

t = 4°C

There will be a rise of  4 degree in the temperature of ice.  

 

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Identify two size dependent properties and two size independent properties of an iron nail
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6 0
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Read 2 more answers
Two equally charged, 2.098 g spheres are placed with 3.338 cm between their centers. When released, each begins to accelerate at
photoshop1234 [79]

Answer:

2.64\times 10^{-7} C

Explanation:

There are two spheres name 1 and 2 and they posses the same charge, which is +q.

And they have equal mass which is 2.098 g.

The distance between these two spheres is, r=3.338 cm.

And the acceleration of each sphere is, a=269.429 m/s^{2}.

Now the coulumbian force experience by 1 sphere due to 2 sphere,

F_{21} =\frac{q^{2} }{4\pi\epsilon_{0} r^{2}  }.

And also the newton force will occur due to this force,

F_{21}=ma.

Now equate the above two values of force will get,

\frac{q^{2} }{4\pi\epsilon_{0} r^{2}  } =ma

Further solve this,

q^{2}=ma4\pi  \epsilon_{0} r^{2}.

Substitute all the known variables in above equation,

q^{2}=(2.098\times 10^{-3} )(269.429)(4(3.14))(8.85\times 10^{-12})(3.338\times 10^{-2}).

q=2.64\times 10^{-7} C.

6 0
3 years ago
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