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Marta_Voda [28]

3 years ago

12

A 200g of iron at 120 degrees and a 150 g piece of copper at -50 degrees are dropped into an insulated beaker containing 300 g o

f ethyl alcohol at 20 degrees. What is the final temperature?

1 answer:

kodGreya [7K]3 years ago

4 0

Answer:

T = 15.03°C

Explanation:

given data:

copper specific heat = Sc = 0.385 J/g °C

iron specific iron = Si = 0.450 J/g °C

specific heat of ethanol = Se = 2.46 J/g °C

net heat loss is equal to zero

$(m*S*\Delta T)_{copper} +(m*S*\Delta T)_ {iron} +(m*S*\Delta T)_ {ethanol} = 0$

$150*0.385 *( T - (-50)) + 200*0.450*(T - 120) + 300*2.46 * (T -20) = 0$

$57.75( T - (-50)) + 0.90(T - 120) +738(T -20) = 0$

$57.75T + 2887.5 + 0.90T - 108 + 738T - 14760 = 0$

$57.75T + 0.90T+738T = - 2887.5 + 108+14760$

$796.65T= 11980.5$

T = 15.03°C

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$\large\displaystyle\text{$\begin{gathered}\sf \bf{A) \ Calculate \ the \ average \ speed: } \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf 402.33 \ m*8 \ laps = 3218.64 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf d=3218.64 \ m \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf t=750 \ s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf V=\frac{d}{t} \ \ \ \ \ \ V= \frac{3218.64 \ m }{750 \ s} \end{gathered}$}\\\\\\\large\displaystyle\text{$\begin{gathered}\sf V=4.29 \ m/s \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf \bf{B) \ Calculate \ the \ average \ speed \ in \ m/s} \end{gathered}$}$

$\large\displaystyle\text{$\begin{gathered}\sf V=402.33 \ m \end{gathered}$}$

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Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i

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Answer:

The final speed for the mass 2m is $v_{2y}=-1,51\ v_{i}$ and the final speed for the mass 9m is $v_{1f} =0,85\ v_{i}$ .

The angle at which the particle 9m is scattered is $\theta = -66,68^{o}$ with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

$\vec{p}=m\vec{v}$

$p_{xi} =p_{xf}$

In the x axis before the collision we have

$p_{xi}=9m\ v_{i} - 2m\ v_{i}$

and after the collision we have that

$p_{xf} =9m\ v_{1x}$

In the y axis before the collision $p_{yi} =0$

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$p_{yf} =9m\ v_{1y} - 2m\ v_{2y}$

so

$p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}$

then

$p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}$

<u>Conservation of kinetic energy:</u>

$\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}$

so

$\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}$

Putting in one side of the equation each speed we get

$\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}$

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the speed of the 9m particle:

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the magnitude of the final speed of the particle 9m is

$v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }$

$v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}$

The tangent that the speed of the particle 9m makes with the -y axis is

$tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}$

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$\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}$

As a vector the speed of the particle 2m is:

$\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}$

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