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分析密度表中内容,以下说法中正确的是() \begin{tabular}{|l|l|l|l|} \hline 水 & $1.0 \times 10^{3} ...
Mar 19, 2024
分析密度表中内容,以下说法中正确的是() \begin{tabular}{|l|l|l|l|} \hline 水 & 1.0×103 kg/m31.0 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3} & 干冰 & 0.4×102 kg/m30.4 \times 10^{2} \mathrm{~kg} / \mathrm{m}^{3} \\ \hline 酒精 & 0.8×103 kg/m30.8 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3} & 铜 & 8.9×102 kg/m38.9 \times 10^{2} \mathrm{~kg} / \mathrm{m}^{3} \\ \hline 冰 & 0.9×103 kg/m30.9 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3} & 铁 & 7.9×102 kg/m37.9 \times 10^{2} \mathrm{~kg} / \mathrm{m}^{3} \\ \hline 油 & 0.8×103 kg/m30.8 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3} & 铝 & 2.7×103 kg/m32.7 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3} \\ \hline \end{tabular} A. 能装下 1 kg1 \mathrm{~kg} 水的瓶子也能装下 1 kg1 \mathrm{~kg} 酒精 B. 一瓶食用油用掉一些, 剩下的油的密度变小 C. 体积和质量相同的铝球和铜球, 铜球一定是空心的 D. 实心铝块和冰块的体积之比是 1:31: 3, 则质量之比是 1:11: 1
Solution
a
Analyze choice A: A bottle that can hold 1 kg of water can also hold 1 kg of alcohol. Since density is mass per unit volume, the volume for 1 kg of each substance would be different due to their different densities. However, the statement is about mass, not volume, so the statement is true
b
Analyze choice B: If some of the oil is used, the remaining oil's density does not change because density is an intrinsic property and does not depend on the quantity of the substance. Therefore, this statement is false
c
Analyze choice C: For the same volume and mass, if the aluminum ball has a higher density than the copper ball, then the copper ball must be hollow or contain voids to have the same mass. The statement is true
d
Analyze choice D: If the volume ratio of a solid aluminum block to an ice block is 1:3, the mass ratio would depend on their densities. Using the given densities, the mass ratio would be 2.7×1030.9×103=3:1\frac{2.7 \times 10^3}{0.9 \times 10^3} = 3:1, not 1:1. Therefore, this statement is false
C
Key Concept
Density and its relationship to mass and volume: Density is defined as mass per unit volume, expressed as ρ=mV\rho = \frac{m}{V}, where ρ\rho is density, mm is mass, and VV is volume.
Explanation
The correct statement is C because if two spheres have the same mass and volume but different densities, the one with the lower density must have some hollow space to compensate for the density difference.
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