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makkiz [27]
3 years ago
11

How does heat energy move by convection when a cook heats a pot of water?

Physics
2 answers:
laiz [17]3 years ago
3 0

The correct choice is

B. Particles at the bottom of the water carry heat energy to the top of the water.

when pot of water is heater, the bottom of pot gets heated. the particles of water in contact with the bottom of the pot gets heat through conduction. after getting heat, these particles of water near the bottom, move away towards top and their position is taken by cooler particles from top. that way heat travels

Y_Kistochka [10]3 years ago
3 0
He answer is B: <span>Particles at the bottom of the water carry heat energy to the top of the water 
</span>
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When 1.14 g of octane (molar mass = 114 g/mol) reacts with excess oxygen in a constant volume calorimeter, the temperature of th
balandron [24]

Answer:

Q (reaction) = -69.7 kJ

Explanation:

Octane reacts with oxygen to give carbon dioxide and water.

C₈H₁₈ + 25 O₂ ---> 16 CO₂ +18 H₂O

This reaction is exothermic in nature. Therefore, the energy is released into the atmosphere. This reaction took place in a calorimeter, there the temperature (T) increases by 10 C. The heat capacity of the calorimeter is 6.97 kJ/C

The heat (q) of the reaction is calculated as follows:

Q= -cT, where c is the heat capacity of the calorimeter and T is the increase in temperature

q = -(6.97) x (10) = -69.7kJ

<em>Since the heat capacity is given in kilo -joule per degree Celsius, therefore, the mass of octane is not required </em>

7 0
3 years ago
Energy can come from electricity, but it<br> can also come from water<br> false<br> true
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Answer:

the answer is truth

Explanation:

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Can I have some tips, tricks, and ideas for dropping an egg 12 feet in the air onto a hard surface without it breaking. It HAS t
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A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th
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Answer:

Part a)

v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}

Part b)

v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

Mv_o = M v_1 + m v_2

here we also use angular momentum conservation

so we have

M v_o d = M v_1 d + \beta mL^2 \omega

also we know that the collision is elastic collision so we have

v_o = (v_2 + d\omega) - v_1

so we have

\omega = \frac{v_o + v_1 - v_2}{d}

also we know

M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})

also we know

v_1 = v_o - \frac{m}{M}v_2

so we have

M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})

mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}

now we have

(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}

v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}

Part b)

Now we know that speed of the ball after collision is given as

v_1 = v_o - \frac{m}{M}v_2

so it is given as

v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})

3 0
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4 years ago
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