计算专题01—密度(期末复习课件)八年级物理下学期新教材苏科版
2026-05-22
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资源信息
| 学段 | 初中 |
| 学科 | 物理 |
| 教材版本 | 初中物理苏科版八年级下册 |
| 年级 | 八年级 |
| 章节 | 第六章 物质的物理属性 |
| 类型 | 课件 |
| 知识点 | 质量和密度 |
| 使用场景 | 同步教学-期末 |
| 学年 | 2026-2027 |
| 地区(省份) | 全国 |
| 地区(市) | - |
| 地区(区县) | - |
| 文件格式 | PPTX |
| 文件大小 | 7.63 MB |
| 发布时间 | 2026-05-22 |
| 更新时间 | 2026-05-22 |
| 作者 | 小帖 |
| 品牌系列 | - |
| 审核时间 | 2026-05-22 |
| 下载链接 | https://m.zxxk.com/soft/57987769.html |
| 价格 | 4.00储值(1储值=1元) |
| 来源 | 学科网 |
|---|
摘要:
这是一份苏科版初中物理八年级期末专项复习课件,聚焦密度计算专题,涵盖基础知识(公式、单位换算、易错点)和8大题型(直接计算、物质鉴别、比值问题等),配套例题解析与提高练习,为学生构建系统学习支架。
资料特色融合核心素养,通过首饰鉴别(计算密度判断纯银)夯实物理观念,比值问题推理、空心实心三种判断法提升科学思维,m-V图像分析强化科学探究,结合氧气瓶、体温计等实例增强应用能力,助力学生掌握解题方法,为教师教学提供高效复习资源。八年级学生处于物理计算能力形成关键期,本资料能帮助其巩固密度计算核心方法,提升解决实际问题的能力,为后续学习奠定基础。
内容正文:
期末复习
Calculation Special Topic
期末专项复习
计算专题——密度
苏科版八年级物理(初中)
1
主要题型
题型一 直接运用公式计算
题型二 物质鉴别
题型三 求比值问题
题型四 特殊物质求密度问题
题型五 𝒎−𝑽图像问题
题型六 空心/实心问题
题型七 混合溶液问题
题型八 列方程解题
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一、基础知识
1、公式:
2、公式变形:计算质量:
计算体积:
3、单位换算:质量单位:1t=1000kg、1kg=1000g、1g=1000mg
体积单位:1mL=1cm3=10-6m3、1L=10-3m3=103m3
密度单位:
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3
4、易错点
(1),三个物理量必须是同一物质的同一状态
(2)单位必须相互对应
——,则m——kg、V——m3
——,则m——g、 V——cm3
5、特殊物质的密度
(1)液体:热胀冷缩:热胀——m不变,V变大,则变小
冷缩——m不变,V变小,则变大
(2)气体:密闭的气体容器使用一部分气体后:V不变,m变小,则变小
一、基础知识
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4
题型一 直接运用公式的计算
二、考法分类
例1:现有一新型材料,经过测量该新型材料质量为158g,体积为20cm3,求该新型材料的密度?
3
解题思路:已知条件、3
解答:解:根据新型材料的密度为
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5
二、考法分类
例2:一个空矿泉水瓶体积始终为 500cm³,装满清水时总质量是 550g,空瓶质量是 50g,已知:食用油密度油=0.9g/cm3。
(1)瓶内清水的密度;
(2)装满食用油时油的质量。
解题思路:
(1)求清水的质量
已知瓶子的体积
3
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6
二、考法分类
解题思路:
(2)已知
解答:解(1)清水的质量为m水=m总−m瓶=550g−50g=500 g
水的体积𝑉水=𝑉瓶= 3
再根据瓶内清水的密度
(2)水的体积𝑉油=𝑉瓶= 3
再根据装满食用油时油的质量为
=0.9g/cm3× 3=450g
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7
题型二 鉴别物质
二、考法分类
例:小明妈妈去旅游时买了一件首饰,小明想鉴定一下首饰是否是纯银的进行了以下操作:用天平测出首饰的质量为90g,用量筒测出首饰的体积为10 cm3 (已知)则
(1)首饰是否是纯银的?
(2)若不是纯银,则用纯银制作出同样大小的首饰需要银的质量为?
3
解题思路:
(1)已知首饰的m、V 大小
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8
二、考法分类
解题思路:
(2)V不变
解答:解(1)根据首饰的密度为
则该首饰不是纯银制作的
(2)纯银首饰的质量为
×
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9
二、考法分类
例:现有一容器,能够装入质量为400g的甲液体,若装入乙液体,则可装入乙液体的质量为320g,则甲、乙两种液体的密度之比为?
解题思路:
V相等已知、
题型三 求比值问题
解答:解:根据
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10
二、考法分类
例1:某厂生产的氧气瓶容量为1000cm³,能够装入16g氧气,可供吸氧200次
(1)全新的氧气瓶,瓶内的氧气密度为?
(2)若每次吸氧量相同,使用100次后,剩余氧气的密度为?
解题思路:
(1)已知满瓶时、
(2) 不变
求一次吸入的氧气剩余100次未吸入氧气质量
题型四 特殊物质求密度问题
cm3
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11
二、考法分类
解答:解(1)根据
(2)吸入一次氧气消耗氧气的质量为 =0.08g
吸取100次后瓶内氧气的质量为=8g
剩余氧气的密度为
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12
二、考法分类
例2:有一水银体温计组装厂,常温下在组装体温计时向玻璃泡中装入了质量为0.68g,体积为0.05mL的水银
(1)常温下水银的密度?
(2)若测量体温结束后,水银的体积为0.08mL,则此时水银的密度为?
解题思路:
(1)已知条件、
(2)
𝑽
L
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13
二、考法分类
解答:解(1)根据
(2)测量体温前后,水银的质量不变
此时,水银的体积为=
则测量体温结束后,水银的密度为:
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14
二、考法分类
例:小明想测量一瓶苹果汁的密度,于是他利用天平和量筒测得几组苹果汁和量筒的总质量m及苹果汁体积V,绘制出其m-V图像如图所示
(1)量筒的质量?
(2)苹果汁的密度
题型五图像问题
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15
二、考法分类
解题思路:
(1)()
苹果汁密度不变
()
(2)
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16
二、考法分类
解答:解(1)由图像可知,当时3
当时3
根据
代入数据:
解得量筒的质量为=16g
(2)由题(1)可得当苹果汁的体积为 3时
苹果汁的质量为= =37g =21g
则苹果汁的密度为= =1.05
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17
二、考法分类
例:现有一个铁球,质量为790g,体积为200cm³,已知铁的密度ρ铁=7.9×10³kg/m³(1)判断该铁球是实心还是空心的?
(2)若为空心,求空心部分的体积?
题型六 空心/实心问题
解题思路:
方法一:比较密度法
(1)已知条件的实际
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18
二、考法分类
解答:解(1)根据
铁球的实际密度为
则该铁球是空心的
解题思路:
方法二:比较质量法
(1)假设的
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19
二、考法分类
解答:解(1)假设铁球是实心的,根据
实心铁球的质量为
=1580g
则该铁球是空心的
解题思路:
方法三:比较体积法
(1)假设的
(2)实心部分体积
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20
二、考法分类
解答:解(1)假设铁球是实心的,根据
实心铁球的体积为V实
=100
则该铁球是空心的
(2)空心部分的体积为
=200 100
=100
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21
二、考法分类
例1:有一个质量为232g的铜铝合金,其中含铝54g,求合金的密度?(已知铜的密度,铝的密度)
题型七问题
解题思路:
已知
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22
二、考法分类
解答:解:铜的质量
根据铜的体积= =20
铝的体积= =20
所以,合金的体积为𝑉合=𝑉铝+𝑉铜
= 20
则合金的密度为 5.8
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23
二、考法分类
例2:酒的度数表示每100mL酒中所含有酒精的毫升数。某工厂制作的一种白酒,瓶上标有450mL,45度的字样(不考虑酒精和水混合时体积的变化。,)
(1)这瓶白酒中酒精的质量是多少?
(2)这瓶白酒中水的质量是多少?
(3)这瓶白酒的密度是多少?
=450mL=450cm3
每100mL白酒中含有45mL的酒精
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24
二、考法分类
解题思路:
(1)计算一瓶白酒中酒精的体积
(2)计算一瓶酒中水的体积
(3)计算白酒的质量
解答:解(1)由题意可知450nL白酒中所含酒精的体积为
cm3
根据酒精的质量
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25
二、考法分类
解答:解(2)白酒中水的体积为
=450mL 202.5m=247.5mL=247.5c
白酒中水的质量为
(3)白酒的质量
=162 =409.5
白酒的密度为=
=
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26
二、考法分类
例:一辆货车的最大载重量为40t,货箱的容积为50 m3,为了使货箱的空间不浪费,现在采取钢材和板材混装的方式装车,则钢材最多装入多少?(已知钢材的密度为,板材的密度为)
题型八 列方程解题
解题思路:
已知最大载重量
←联立解
已知货车容积 50
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27
二、考法分类
解答:解:设钢材的体积为,则板材的体积为50
由题意可知,
根据,
联立上式:+ (50
40 ,
解得
则钢材的最大质量
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28
三、提高练习
1、一个铁球质量790g,体积200cm3,铁=7.9g/cm3
(1)判断铁球实心还是空心?
(2)空心部分灌满某种液体后总质量870g,求该液体密度?
解答(1)空心
(2)0.8g/cm3
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29
三、提高练习
2、甲乙两个实心圆柱体,底面积之比,高度之比,质量之比,求甲乙物质密度之比?
解答:
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30
三、提高练习
3、将质量为178g的铜块和270g铝块熔铸成实心合金块(不计体积损耗),铜=8.9g/cm3,铝=2.7g/cm3
求:合金块的平均密度
解答:3.73g/cm3
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31
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