- The heat of vaporization of ethyl alcohol Lv is about 200 cal/g. Show that if 4 kg of this refrigerant were allowed to vaporize in a refrigerator, it could freeze 10 kg of 0°C water to ice.
- From greatest to least, rank the frequency of radiation of these emitters of radiant energy: (a) red-hot star, (b) bluehot star, and (c) the Sun.
The Correct Answer and Explanation is :
Problem 1: Heat of Vaporization of Ethyl Alcohol
To show that 4 kg of ethyl alcohol (Lv ≈ 200 cal/g) could freeze 10 kg of 0°C water into ice, let’s go through the steps systematically.
- Heat required to freeze 10 kg of water:
The latent heat of fusion for water (Lf) is approximately 80 cal/g. The amount of heat needed to freeze 10 kg (or 10,000 grams) of water at 0°C is: [
Q = m \times L_f
]
Where:
- ( m = 10,000 ) g (the mass of water)
- ( L_f = 80 ) cal/g (latent heat of fusion) Therefore:
[
Q = 10,000 \, \text{g} \times 80 \, \text{cal/g} = 800,000 \, \text{cal}
]
So, it requires 800,000 cal to freeze 10 kg of water.
- Heat released by vaporizing 4 kg of ethyl alcohol:
The heat of vaporization for ethyl alcohol (Lv) is given as 200 cal/g. The amount of heat released when 4 kg (or 4,000 grams) of ethyl alcohol vaporizes is: [
Q = m \times L_v
]
Where:
- ( m = 4,000 ) g (the mass of ethyl alcohol)
- ( L_v = 200 ) cal/g (latent heat of vaporization) Therefore:
[
Q = 4,000 \, \text{g} \times 200 \, \text{cal/g} = 800,000 \, \text{cal}
]
This means 4 kg of ethyl alcohol would release 800,000 cal of energy during vaporization, which exactly matches the amount of energy required to freeze 10 kg of water.
So, we can conclude that if 4 kg of ethyl alcohol were allowed to vaporize in a refrigerator, it could indeed freeze 10 kg of 0°C water into ice.
Problem 2: Ranking the Frequency of Radiation from Different Emitters
To rank the frequency of radiation from different emitters, we need to understand the relationship between temperature and the frequency of emitted radiation. The key law that governs this relationship is Wien’s Displacement Law, which states that:
[
\lambda_{\text{max}} = \frac{b}{T}
]
Where:
- ( \lambda_{\text{max}} ) is the wavelength at which the intensity of radiation is maximum.
- ( b ) is Wien’s constant (approximately ( 2.9 \times 10^{-3} \, \text{m} \cdot \text{K} )).
- ( T ) is the absolute temperature of the emitter in Kelvin.
From Wien’s law, the higher the temperature, the shorter the wavelength at which the maximum radiation occurs. Since the frequency (( f )) of radiation is inversely related to the wavelength, we have:
[
f = \frac{c}{\lambda}
]
Where:
- ( f ) is the frequency of the radiation.
- ( c ) is the speed of light in a vacuum (approximately ( 3.0 \times 10^8 \, \text{m/s} )).
Thus, a higher temperature corresponds to a higher frequency of emitted radiation.
Now, let’s rank the three emitters:
- Blue-hot star: A blue-hot star is hotter than a red-hot star, emitting radiation at shorter wavelengths. According to Wien’s Law, the hotter the object, the higher the frequency of radiation. Therefore, the blue-hot star will emit radiation at the highest frequency.
- Sun: The Sun has a temperature of around 5,500 K. Its radiation peaks in the visible spectrum, typically around yellow or green light. Its frequency is lower than that of a blue-hot star, but higher than that of a red-hot star.
- Red-hot star: A red-hot star has a lower temperature than both the blue-hot star and the Sun. Therefore, it emits radiation at longer wavelengths, resulting in a lower frequency of radiation.
Ranking from greatest to least frequency of radiation:
- Blue-hot star
- Sun
- Red-hot star
Explanation:
The frequency of radiation emitted by an object is directly related to its temperature. The higher the temperature, the shorter the wavelength and the higher the frequency of the emitted radiation. A blue-hot star, being the hottest of the three, emits radiation at the highest frequency, followed by the Sun, which has a moderate temperature, and finally the red-hot star, which is the coolest of the three and emits radiation at the lowest frequency.