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Stella [2.4K]
3 years ago
9

Heat in a substance is an example of ________ energy. A. phase B. kinetic C. potential D. evaporative

Chemistry
2 answers:
lisov135 [29]3 years ago
8 0
The correct answer that would best complete the given statement above is option B. Heat in a substance is an example of a kinetic energy. Kinetic energy is a kind of energy that is in motion. We can feel the kinetic energy of heat through their temperatures because heat is being transferred. 
Marina86 [1]3 years ago
8 0

Answer: Option (B) is the correct answer.

Explanation:

When there will be flow of energy from one substance to another then it is defined as the heat.

Since, heat is responsible for the movement of particles of a substance. This means that more if the heat more will be the kinetic energy as kinetic energy is the energy obtained by particles of a substance due to their motion.

Also, kinetic energy is directly proportional to temperature.

                     K.E = \frac{3}{2}kT

So, more is the temperature of an object more will be heat transferred by it. Therefore, more will be its kinetic energy.

Thus, we can conclude that heat in a substance is an example of kinetic energy.

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Answer:

This is google's answer for the last question

Explanation:

The kinetic energy increases as the particles move faster. The potential energy increases as the particles move farther apart. How are thermal energy and temperature related? When the temperature of an object increases, the average kinetic energy of its particles increases.

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3 years ago
Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t
Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

Q_1=m\times \Delta H_{fusion}

where,

Q_1 = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

\Delta H_{fusion} = enthalpy change for fusion = 3.35\times 10^5J/Kg

Now put all the given values in Q_1, we get:

Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J

<u>For process 2 :</u>

Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})

where,

Q_2 = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

c_{p,l} = specific heat of liquid water = 4186J/Kg^oC

T_1 = initial temperature = 0^oC

T_2 = final temperature = 100^oC

Now put all the given values in Q_2, we get:

Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC

Q_2=4.186\times 10^5J

From this we conclude that, Q_1 that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

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