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3 years ago

Explain how heat transfer is occurring through convection, radiation AND conduction in this picture.

Physics

1 answer:

frosja888 [35]3 years ago

4 0

A cup of tea/coffee transfers heat via conduction, convection and radiation.

Conduction:

Heat transfer from a coffee to the spoon is in direct contact
Inner part of the tea cup against hot liquid will transfer heat to the outer part and saucer.

Convection:

Movement of molecules within the coffee. Hot molecules of the tea at the middle of the drink will be eventually make their way to the outer part of the drink near the tea cup, where they will loose their heat.

Radiation:

Thermal radiation is the loss of heat from a solid object in to atmosphere via the release of electromagnetic radiation. So hot outer edge of the teacup loss the energy to its surroundings via thermal radiation.

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The compound PCl5 decomposes into Cl2 and PCl3. The equilibrium of PCl5(g) Cl2(g) + PCl3(g) has a Keq of 2.24 x 10-2 at 327°C. W

nordsb [41]

Take note of the reaction formula which is PCl5=Cl2+PCl3.
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3 years ago

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torisob [31]

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3 years ago

Qual a capacidade térmica de um objeto que, ao receber 10000 cal de energia, tem sua temperatura elevada de 25°C para 75°C?

Stolb23 [73]

Answer:

$200 cal/^{\circ}C$

Explanation:

When heat energy is supplied to an object, the temperature of the object increases according to the equation:

$Q=C\Delta T$

where

Q is the heat supplied

C is the heat capacity of the object

$\Delta T$ is the change in temperature

In this problem we have:

$Q=10,000 cal$ is the energy supplied

$\Delta T=75C-25C=50C$ is the change in temperature of the object

Therefore, the heat capacity of the object is:

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3 years ago

A 2 kg ball is moving 3 m/s when it starts rolling up a hill.

AURORKA [14]

Answer:

the height reached is = 0.458 [m]

Explanation:

We need to make a sketch of the ball and see the location of the reference point where the potential energy is zero. But the kinetic energy will be defined by the following expression:

$Ek=\frac{1}{2} *m*v^{2} \\where:Ek= kinetic energy [J]\\m = mass of the ball [kg]\\v = velocity of the ball [m/s]$

Replacing the values on the equation we have:

$Ek=\frac{1}{2}*(2)*(3^{2} )\\ Ek=9[J]\\$

This kinetic energy will be transformed in potential energy in the moment when the ball starts to rolling up. Therefore the maximum height reached by the ball depends of the initial velocity given to the ball.

$Ek=Ep\\where\\Ep=potential energy [J]\\Ep=m*g*h\\where\\g=gravity = 9.81[m/s^2]\\h=height reached [m]\\$

Now we have:

$h=\frac{Ep}{m*g} \\h=\frac{9}{2*9.81} \\\\h=0.45 [m]$

In that moment when the ball reach the 0.45 [m] the potencial energy will be maximum and equal to the kinetic energy when the ball has a velocity of 3[m/s]

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3 years ago

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Aliun [14]

Answer:

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