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Thermal energy is just randomized kinetic energy on a microscopic scale. To gain some insight into this, consider one liter ( \(1 \mathrm{~kg}\) ) of water, and figure out how much energy it would take to heat it from absolute zero temperature \(^{37}\) to \(300 \mathrm{~K}\) assuming that the definition of the calorie (Sec. 5.5; p.73) applies across this entire range. If this same amount of energy went into kinetic energy- hurling the water across the room-what would the corresponding velocity be?

Short Answer

Expert verified
Answer: It requires 1,255,800 J of energy to heat 1 kg of water from 0 K to 300 K. If this energy were converted to kinetic energy, the water would have a velocity of about 1584.17 m/s.

Step by step solution

01

Calculate the energy required to heat the water

First, we need to calculate the amount of energy required to heat the water from 0 K to 300 K. We can use the specific heat formula for this calculation: \(Q = mc\Delta T\) where: - \(Q\) is the energy required - \(m\) is the mass of the water (1 kg) - \(c\) is the specific heat of water (1 cal/g.K \(\approx\) 4.186 J/g.K) - \(\Delta T\) is the temperature change (300 K - 0 K = 300 K) Note that 1 kg = 1000 g for our conversion.
02

Calculate the energy in joules

Now that we have the specific heat formula, we can plug in the values and convert the result from calories to joules: \(Q = (1 \,\text{kg})(4.186 \,\text{J/g.K})(300 \,\text{K})\) \(Q = 1000 \,\text{g} \times 4.186 \,\text{J/g.K} \times 300 \,\text{K}\) \(Q = 1255800 \,\text{J}\) So, it takes 1,255,800 J of energy to heat the water from 0 K to 300 K.
03

Calculate the velocity from the kinetic energy

Now, we'll find the corresponding velocity if the same amount of energy were converted to kinetic energy. We can use the kinetic energy formula for this step: \(E_k = \frac{1}{2}mv^2\) where: - \(E_k\) is the kinetic energy (equal to the energy calculated in step 2: 1,255,800 J) - \(m\) is the mass of the water (1 kg) - \(v\) is the velocity of the water Rearranging the formula to solve for the velocity, we get: \(v = \sqrt{\frac{2E_k}{m}}\)
04

Calculate the velocity of the water

Now substitute the values of \(E_k\) and \(m\) into the formula to find the velocity: \(v = \sqrt{\frac{2 \times 1255800 \,\text{J}}{1 \,\text{kg}}}\) \(v = \sqrt{2511600}\) \(v \approx 1584.17 \,\text{m/s}\) So, if the same amount of energy required to heat the water from 0 K to 300 K were to be converted into kinetic energy, the water would have a velocity of approximately 1584.17 m/s.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Specific Heat
Specific heat is the amount of heat per unit mass required to raise the temperature by one degree Celsius. It represents how much energy a substance can absorb before increasing in temperature. For water, its specific heat capacity is quite high, specifically 1 calorie per gram per degree Celsius, equivalent to 4.186 joules per gram per degree Kelvin. This means water can take in a lot of heat before it gets hot.
It's important to understand that the specific heat capacity can also tell us about a substance's ability to store thermal energy. The higher the specific heat, the more energy it can store.
When heating water from absolute zero to a moderate temperature, like 300 K, it's essential to consider that water undergoes a substantial temperature change while absorbing an immense amount of heat. This is why water is an excellent medium for thermal management in various applications.
Kinetic Energy
Kinetic energy refers to the energy an object possesses due to its motion. The faster an object moves, the more kinetic energy it has. In our problem, all the energy used to heat water was eventually converted into kinetic energy.
The mathematical formula for kinetic energy is:
  • \( E_k = \frac{1}{2}mv^2 \)
where \( E_k \) is the kinetic energy, \( m \) is the mass, and \( v \) is the velocity. From this equation, you can see that kinetic energy depends on both the mass and the square of the velocity.
In the exercise, when the heat energy (1,255,800 J) initially used to heat the water was converted entirely into kinetic energy, it resulted in a very high velocity of about 1584.17 m/s for the water. This illustrates how a large amount of energy can result in phenomenal speeds if converted to motion instead of heat.
Energy Conversion
Energy conversion is the process of changing energy from one form to another. In physics, this often involves switching between potential energy, kinetic energy, heat energy, and sometimes even different sectors like electrical energy.
In the context of the exercise, the conversion involves thermal energy being transformed into kinetic energy. This transformation helps illuminate how much energy can be stored as heat and how it translates into the physical movement of mass.
It's through this conversion that we understand the versatile nature of energy. It doesn't disappear but rather changes from one form to another. This concept is crucial when calculating how energy affects the physical states of substances or their movement. Understanding this allows us to efficiently manage energy resources in physical systems, ranging from engines to everyday appliances.

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Most popular questions from this chapter

A hard slap might consist of about \(1 \mathrm{~kg}\) of mass moving at \(10 \mathrm{~m} / \mathrm{s}\). How much kinetic energy is this, and how much warmer would \(10 \mathrm{~g}\) of \(\mathrm{skin}^{33}\) get if the skin has the heat capacity properties of water, as in the definition of a calorie (Sec. 5.5; p. 73 and Sec. \(6.2\); p. 85 are relevant)?

Atmospheric pressure is about \(10^{5} \mathrm{~N} / \mathrm{m}^{2}\), meaning that a \(100,000 \mathrm{~N}\) weight of air-corresponding to a mass of \(10,000 \mathrm{~kg}\) -sits atop very square meter of the ground (at or near sea level). If the air density were constant at \(1.25 \mathrm{~kg} / \mathrm{m}^{3}\) - rather than decreasing with height as it actually does - how high would the atmosphere extend to result in this weight (mass)?

Referring to Figure 12.7, examine performance at \(5 \mathrm{~m} / \mathrm{s}\) and at \(10 \mathrm{~m} / \mathrm{s}\), picking a representative power for each in the middle of the cluster of black points, and assigning a power value from the left-hand axis. What is the ratio of power values you read off the plot, and how does this compare to theoretical expectations for the ratio going like the cube of velocity?

Traveling down the road, you carefully watch a three-bladed wind turbine, determining that it takes two seconds to make a full revolution. Assuming it's operating near the peak of its efficiency curve \(^{41}\) according to Figure \(12.4\), how fast do you infer the wind speed to be if the blade length \(^{42}\) appears to be \(15 \mathrm{~m}\) long?

The largest wind turbines have rotor diameters \(^{40}\) around \(150 \mathrm{~m}\). Using a sensible efficiency of \(50 \%\), what power does such a jumbo turbine deliver at a maximum design wind speed of \(13 \mathrm{~m} / \mathrm{s} ?\)

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