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Chapter 12: Problem 5

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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Most popular questions from this chapter

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?

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)?

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 \(13 \mathrm{~m} / \mathrm{s} ?\)

Considering that wind turbines are rated for the maximumtolerable wind speed around \(12 \mathrm{~m} / \mathrm{s}\), and tend to operate at about \(30 \%\) capacity factor, how much average power \(^{44}\) would a \(100 \mathrm{~m}\) diameter turbine operating at \(45 \%\) efficiency be expected to produce?

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?
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