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Airida [17]
3 years ago
9

550 g of water at 105°C is poured into an 855 g aluminum container with an initial temperature of 11°C. The specific heat of alu

minum is 900 J/(kg K). How much heat flows from the water to the aluminum, assuming no heat is exchanged with the surroundings?
Physics
1 answer:
nexus9112 [7]3 years ago
5 0

Answer:

54.22 kJ

Explanation:

In this case, we need to calculate the heat. The expression to use is:

Q = m * C * ΔT

Now, with the specific heat of water (4186 J/ kg K), we can calculate the temperature in which this occurs.

So:

Q = 0.550 * 4186 * (105 - T)

Q = 2302.3 (105 - T)

Q = 241,741.5 - 2302.3T

Now with the Aluminium:

Q = 0.855 * 900 * (T - 11)

Q = 769.5T - 8464.5

Now, with both equations, we solve for the final temperature:

769.5T - 8464.5 = 241,741.5 - 2302.3T

(2302.3 + 769.5)T = 241,741.5 + 8464.5

3071.8T = 250,206

T = 81.45 K

This is the temperature which the change occurs. Now, let's determine the amount of heat from water to Al:

Q = 241,741.5 - 2302.3(81.45)

Q = 54,219.17 J or simply 54.22 kJ.

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mass weighing 16 pounds stretches a spring 8 3 feet. The mass is initially released from rest from a point 2 feet below the equi
valina [46]

Answer:

The answer is

"x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))".

Explanation:

Taking into consideration a volume weight = 16 pounds originally extends a springs \frac{8}{3} feet but is extracted to resting at 2 feet beneath balance position.

The mass value is =

W=mg\\m=\frac{w}{g}\\m=\frac{16}{32}\\m= \frac{1}{2} slug\\

The source of the hooks law is stable,

16= \frac{8}{3} k \\\\8k=16 \times 3 \\\\k=16\times \frac{3}{8} \\\\k=6 \frac{lb}{ft}\\\\

Number \frac{1}{2}  times the immediate speed, i.e .. Damping force

\frac{1}{2} \frac{d^2 x}{dt^2} = -6x-\frac{1}{2}\frac{dx}{dt}+10 \cos 3t \\\\\frac{1}{2}  \frac{d^2 x}{dt^2}+ \frac{1}{2}\frac{dx}{dt}+6x =10 \cos 3t \\ \\\frac{d^2 x}{dt^2} +\frac{dx}{dt}+12x=20\cos 3t \\\\

The m^2+m+12=0 and m is an auxiliary equation,

m=\frac{-1 \pm \sqrt{1-4(12)}}{2}\\\\m=\frac{-1 \pm \sqrt{47i}}{2}\\\\\ m1= \frac{-1 + \sqrt{47i}}{2} \ \ \ \ or\ \ \ \ \  m2 =\frac{-1 - \sqrt{47i}}{2}

Therefore, additional feature

x_c (t) = e^{\frac{-t}{2}}[C_1 \cos \frac{\sqrt{47}}{2}t+ C_2 \sin \frac{\sqrt{47}}{2}t]

Use the form of uncertain coefficients to find a particular solution.  

Assume that solution equation,

x_p = Acos(3t)+B sin(3t) \\x_p'= -3A sin (3t) + 3B cos (3t)\\x_p}^{n= -9 Acos(3t) -9B sin (3t)\\

These values are replaced by equation ( 1):

\frac{d^2x}{dt}+\frac{dx}{dt}+ 12x=20 \cos(3t) -9 Acos(3t) -9B sin (3t) -3Asin(3t)+3B cos (3t) + 12A cos (3t) + 12B sin (3t)\\\\3Acos 3t + 3B sin 3t - 3Asin 3t + 3B cos 3t= 20cos(3t)\\(3A+3B)cos3t -(3A-3B)sin3t = 20 cos (3t)\\

Going to compare cos3 t and sin 3 t coefficients from both sides,  

The cost3 t is 3A + 3B= 20 coefficients  

The sin 3 t is 3B -3A = 0 coefficient  

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Replace the very first equation with the meaning,

3B -3A=O\\3(\frac{10}{3})-3A =0\\A= \frac{10}{3}\\

equation is

x_p\\\\\frac{10}{3} cos (3 t) + \frac{10}{3} sin (3t)

The ultimate plan for both the equation is therefore

x(t)= e^\frac{-t}{2} (c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)

Initially, the volume of rest x(0)=2 and x'(0) is extracted by rest i.e.  

Throughout the general solution, replace initial state x(0) = 2,

Replace x'(0)=0 with a general solution in the initial condition,

x(t)= e^\frac{-t}{2} [(c_1 cos \frac{\sqrt{47}}{2}t)+c_2\sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}\cos (3t)+\frac{10}{3}\sin (3t)]\\\\

x(t)= e^\frac{-t}{2} [(-\frac{\sqrt{47}}{2}c_1\sin\frac{\sqrt{47}}{2}t)+ (\frac{\sqrt{47}}{2}c_2\cos\frac{\sqrt{47}}{2}t)+c_2\cos\frac{\sqrt{47}}{2}t)  +c_1\cos\frac{\sqrt{47}}{2}t +c_2\sin\frac{\sqrt{47}}{2}t + \frac{-1}{2}e^{\frac{-t}{2}} -10 sin(3t)+10 cos(3t) \\\\

c_2=\frac{-64\sqrt{47}}{141}

x(t)= e^\frac{-t}{2}((\frac{-4}{3})\cos\frac{\sqrt{47}}{2}t- \frac{-64\sqrt{47}}{141} \sin\frac{\sqrt{47}}{2}t)+\frac{10}{3}(\cos(3t)+ \sin (3t))

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The earth has a vertical electric field at the surface, pointing down, that averages 119 N/C . This field is maintained by vario
velikii [3]

Answer:

q=5.37*10^{5}C

Explanation:

If we assume that the Earth is a spherical conductor, according to  Gauss's Law, the electric field is given by:

E=\frac{kq}{r^2}

Here k is the Coulomb constant, the excess charge on the Earth's surface and r its radius. Solving for q:

q=\frac{Er^2}{k}\\q=\frac{119\frac{N}{C}(6.371*10^6m)^2}{8.99\frac{N\cdot m^2}{C^2}}\\q=5.37*10^{5}C

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Answer:

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Nutka1998 [239]

You want to use PEMDAS to solve this equation.

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Answer:

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Explanation:

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8 0
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