2026年浙教版适配高三物理暑假衔接卷声光热综合应用标准试卷第262套（含答案解析与可打印作答区）

高三物理暑假衔接卷·声光热综合应用本卷总分120分考试时间120分钟高三物理暑假衔接综合应用检测卷（声、光、热）学校：______________班级：______________姓名：______________考号：______________考试时间：120分钟满分：120分题型单项选择填空与基础实验与材料综合解答分值36分32分32分20分答题说明：1.全卷共四大题，28小题。请先通读题目，再按要求在指定位置作答。2.计算题应写出必要公式、代入过程、单位和结论。3.选择题只有一个最佳答案，多选、错选或不选均不得分。4.书写应清楚，作图和表格数据应与题意一致。诚信提示：独立完成检测，保持卷面整洁，不在密封线外书写与答案无关内容。选择题答题栏（请将所选字母写入表格）：题号123456789101112答案一、单项选择题（本大题共12小题，每小题3分，共36分）1.（3分）某同学用频率为680Hz的音叉在空气中发声，室温下空气中声速取340m/s。该声波在空气中的波长最接近（）。A.0.50mB.1.00mC.2.00mD.5.00m2.（3分）一束单色光由空气斜射入玻璃时，下列关于光的物理量变化的判断正确的是（）。A.频率变小，波长变短B.频率变大，波速变小C.频率不变，波速变小D.频率不变，波长变长3.（3分）两根长度和横截面积相同的金属棒分别连接在同一热源和同一冷源之间，若甲棒单位时间传递的热量约为乙棒的3倍，则最合理的解释是（）。A.甲棒的温度一定比乙棒高3倍B.甲棒材料的导热本领约为乙棒的3倍C.甲棒的比热容一定比乙棒小3倍D.甲棒两端温差一定比乙棒大3倍4.（3分）在同一位置测得某机器工作时的声强级由50dB升至60dB，则声强约变为原来的（）。A.2倍B.6倍C.10倍D.100倍5.（3分）焦距为10cm的薄凸透镜前30cm处放置一支小蜡烛，光屏上得到清晰像。像的性质为（）。A.倒立、缩小的实像B.倒立、放大的实像C.正立、放大的虚像D.正立、等大的虚像6.（3分）关于物体热辐射，下列说法正确的是（）。A.只有温度高于100℃的物体才辐射电磁波B.物体温度升高时，辐射峰值对应波长一般变长C.黑色物体只吸收光，不会辐射能量D.物体温度升高时，总辐射功率一般增大7.（3分）一根两端固定的弦线上形成稳定驻波。相邻两个波节之间的距离为（）。A.四分之一波长B.半个波长C.一个波长D.两个波长8.（3分）光纤能将光信号沿弯曲路径传送，主要依靠的条件是（）。A.纤芯折射率小于包层折射率，且光线反复折射B.纤芯折射率等于包层折射率，且光线保持直线传播C.包层吸收光后再向前发光D.纤芯折射率大于包层折射率，且入射角满足全反射条件9.（3分）温度计放入水中一段时间后读数稳定。读数稳定所反映的主要物理思想是（）。A.水对温度计做功为零B.温度计与水达到热平衡C.温度计的内能为零D.温度计的质量保持不变10.（3分）救护车鸣笛向静止的观察者靠近时，观察者听到的音调比车上声源发出的音调高。形成这一现象的原因是（）。A.观察者接收到的声波频率增大B.声波在空气中的传播速度增大C.声波的振幅在传播中增大D.声源发出的频率随距离自动增大11.（3分）在杨氏双缝干涉实验中，其他条件不变，仅将双缝间距增大为原来的2倍，则屏上相邻亮纹间距将（）。A.增大为原来的2倍B.减小为原来的1/2C.保持不变D.变为原来的4倍12.（3分）一定质量的理想气体保持体积不变并缓慢升温。下列说法正确的是（）。A.气体压强减小，分子平均动能增大B.气体压强不变，分子平均动能减小C.气体压强增大，内能增大D.气体对外做功，内能不变二、填空与基础题（本大题共8小题，每小题4分，共32分）13.（4分）岸边同学对着湖对岸山崖大喊，1.20s后听到回声。取空气中声速340m/s，则同学到山崖的距离为__________m；若气温升高，声速变大，测得的回声时间将__________。答：______________________________________________________________________________________________________________________________________________________________________________14.（4分）焦距为12cm的薄凸透镜成像实验中，物距为24cm时，像距为__________cm，像的大小与物体相比为__________。答：______________________________________________________________________________________________________________________________________________________________________________15.（4分）真空中波长为600nm的单色光进入折射率为1.50的玻璃。光在玻璃中的速度为__________m/s，玻璃中的波长为__________nm。答：______________________________________________________________________________________________________________________________________________________________________________16.（4分）质量为0.20kg的水由20℃升高到45℃，水的比热容取4.2×10^3J/(kg·℃)。水吸收的热量为__________J；若电热器向外提供的电能为3.0×10^4J，则本过程的有效加热比例为__________。答：______________________________________________________________________________________________________________________________________________________________________________17.（4分）空气中某点的声强级为60dB，取基准声强I0=1.0×10^-12W/m^2。该点声强为__________W/m^2；若两台完全相同声源在该点相干效应不明显，声强约加倍，声强级增加约__________dB。答：______________________________________________________________________________________________________________________________________________________________________________18.（4分）一根1.000m长的钢尺从10℃升高到60℃，钢的线膨胀系数取1.2×10^-5℃^-1。钢尺伸长量为__________m，即__________mm。答：______________________________________________________________________________________________________________________________________________________________________________19.（4分）一定质量理想气体初态为p1=1.0×10^5Pa、V1=2.0L、T1=300K，末态为V2=3.0L、T2=450K，则末态压强p2=__________Pa；判断依据是__________。答：______________________________________________________________________________________________________________________________________________________________________________20.（4分）一束光照到平面镜上，入射角为40°，反射角为__________；若该光再由空气斜射入水中，折射角一般__________入射角。答：______________________________________________________________________________________________________________________________________________________________________________三、实验与材料分析题（本大题共4小题，每小题8分，共32分）21.（8分）某实验小组用两个相同传声器测量空气中的声速。两传声器沿声波传播方向相距0.85m，脉冲声先后到达两传声器的时间差为2.50ms。请回答：（1）根据数据求声速；（2）若两传声器连线与声波传播方向并不完全重合，测得声速偏大还是偏小？（3）写出一种减小随机误差的方法。答：__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________22.（8分）在探究凸透镜成像规律时，某组记录如下表。已知烛焰、凸透镜和光屏中心已调整在同一高度。序号物距u/cm像距v/cm像的性质甲30.015.0倒立、缩小、实像乙20.0待求倒立、等大、实像丙15.030.0倒立、放大、实像请回答：（1）由甲组数据求该凸透镜焦距；（2）填写乙组像距；（3）若实验中光屏上像偏高，应怎样调节光屏或烛焰；（4）说明为什么物距小于焦距时不能在光屏上接到实像。答：__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________23.（8分）为测量水的比热容，某组用电热器给烧杯中0.25kg水加热。电热器功率为84W，通电100s，水温由22.0℃升至30.0℃。温度计读数稳定后记录，实验中不搅拌时水面附近和杯底温度有明显差别。请回答：（1）按电能全部被水吸收估算水的比热容；（2）不搅拌会使温度读数怎样影响计算结果；（3）写出两条改进措施。答：__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________24.（8分）阅读材料并作答。冬季温室常采用透明覆盖层。太阳短波辐射较容易透过覆盖层，被土壤和植物吸收后转化为内能；土壤再向外发出较长波长的红外辐射，覆盖层对部分红外辐射透过较弱。同时，封闭空间减弱了室内外空气的直接交换。夜晚可在温室内放置装水桶，白天水吸热升温，夜晚水向周围放热，从而减小温度波动。请回答：（1）温室保温涉及哪两种主要传热方式的调控；（2）为什么水桶有助于夜间保温；（3）若只增加覆盖层厚度但不处理通风缝隙，保温效果可能仍不理想，原因是什么；（4）从光学和热学角度各提出一条改进建议。答：__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________四、综合解答题（本大题共4小题，每小题5分，共20分）25.（5分）超声测距仪发出一束超声脉冲，遇到墙面后返回。发出脉冲到接收到回波共用0.120s，取空气中声速340m/s。求测距仪到墙面的距离；若环境温度升高而仪器仍按340m/s计算，测距结果会怎样偏离真实值？答：__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________26.（5分）一束单色光从空气以30°入射角射入折射率为1.50的透明平板。求折射角的正弦值；说明光进入平板后频率、波速、波长分别怎样变化。答：__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________27.（5分）某热机每个循环从高温热源吸收600J热量，向低温热源放出420J热量。求每个循环对外做功、热机效率；若每秒完成20个循环，求输出功率。答：__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________28.（5分）一台简易太阳能灶用凹面反射面把阳光会聚到黑色水壶上。反射面有效受光面积为0.50m^2，阳光照射功率密度取800W/m^2，有效利用比例为50%。若要把1.0kg水从20℃加热到80℃，水的比热容取4.2×10^3J/(kg·℃)，求至少需要的加热时间；再从光学和热学角度说明提高加热效果的两种做法。答：______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________计算草稿区以下区域供整理公式、单位换算和过程草稿使用，正式评分以各题作答区内容为准。__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________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参考答案与解析说明：以下答案按题号逐题对应。计算结果在合理有效数字范围内均可给分；解答题应同时看过程、单位与结论。一、单项选择题题号123456789101112答案ACBCADBDBABC1.A。声波满足v=λf，λ=v/f=340/680=0.50m。B、C、D均由计算关系不符得到，未体现频率与波长的反比关系。2.C。光从空气进入玻璃时，介质改变但光源振动频率不变；玻璃中波速v=c/n，小于空气中近似光速，波长随波速减小。3.B。在长度、横截面积和两端温差相同条件下，单位时间传热多少主要由材料导热本领决定。比热容描述升温难易，不直接决定稳态导热速率。4.C。声强级每增加10dB，对应声强变为10倍。50dB到60dB正好增加10dB，因此声强约为原来的10倍。5.A。物距30cm大于2倍焦距20cm，凸透镜成倒立、缩小的实像，像距位于f与2f之间。6.D。任何温度高于绝对零度的物体都存在热辐射；温度升高时总辐射功率一般增大，辐射峰值向短波方向移动。7.B。驻波中相邻波节之间为半个波长，相邻波腹之间也是半个波长，波节与相邻波腹相距四分之一波长。8.D。光纤传光依靠纤芯相对包层折射率较大，使进入纤芯的光在芯包界面满足全反射条件并沿纤芯传播。9.B。温度计读数稳定说明温度计与水之间没有净热量交换，二者达到热平衡，其共同温度由温度计读出。10.A。声源向观察者靠近时，观察者在单位时间内接收到更多波峰，接收频率增大，听到音调升高；空气中声速主要由介质状态决定。11.B。双缝干涉相邻亮纹间距Δx=λL/d。波长和屏距不变，缝距d增大为2倍，条纹间距减小为原来的1/2。12.C。定容升温时气体体积不变，对外做功为零；温度升高使分子平均动能增大，理想气体内能增大，压强随温度升高而增大。二、填空与基础题13.204；变短。解析：回声经历往返路程，2d=vt=340×1.20，得d=204m。温度升高时声速增大，同一往返路程所用时间减小。14.24；等大。解析：u=2f=24cm时，凸透镜成像也在2f处，像距v=24cm，成倒立、等大的实像。15.2.0×10^8；400。解析：v=c/n=3.0×10^8/1.50=2.0×10^8m/s；频率不变，λ=λ0/n=600/1.50=400nm。16.2.1×10^4；70%。解析：Q=