2026年浙教版适配专升本物理期中质量检测卷声光热综合应用标准试卷第462套（含答案解析与可打印作答区）

专升本物理期中质量检测卷·声光热综合应用第页2026年浙教版适配专升本物理期中质量检测卷声光热综合应用标准试卷第462套（含答案解析与可打印作答区）学校：________________班级：________________姓名：________________考号：________________考试时间：120分钟满分：120分适用对象：专升本范围：声学、光学、热学综合应用题型结构单项选择36分填空24分实验材料与综合解答60分答题说明与诚信提示1.本卷共28题，满分120分，考试时间120分钟。请在规定区域内作答，保持卷面整洁。2.选择题每题只有一个最佳答案；填空题结果须带必要单位；计算题应写出主要公式、代入过程和结论。3.常用数据如题中未给出，可按g=10N/kg、水的比热容4.2×10³J/(kg·℃)、常温空气声速约340m/s处理。4.独立完成检测，不抄袭、不夹带与考试无关资料。客观题答题栏123456789101112一、单项选择题（本大题共12小题，每小题3分，共36分）1.（3分）一列频率为500Hz的声波在20℃空气中传播，声速取343m/s。关于这列声波，下列说法正确的是（）。A.波长约为0.686mB.振源每秒只振动343次C.声波在真空中也能传播D.声波强弱只由频率决定作答：________2.（3分）同一音叉先后在空气和水中发声，观察者均位于足够近处。忽略能量损失时，下列判断最合理的是（）。A.频率在两种介质中都由音叉决定B.波长在水中一定更短C.声速在水中小于空气中D.音调在水中必然降低作答：________3.（3分）用驻波法测声速时，若相邻两个共振管长读数差为17.2cm，音叉频率为1000Hz，则由实验得到的声速最接近（）。A.172m/sB.344m/sC.688m/sD.1000m/s作答：________4.（3分）一束单色光由空气斜射入玻璃。关于折射光线和频率的变化，下列说法正确的是（）。A.折射光线靠近法线，频率不变B.折射光线远离法线，频率增大C.折射光线方向不变，波长不变D.折射光线靠近法线，频率减小作答：________5.（3分）凸透镜焦距为10cm，物体放在透镜前30cm处。屏上能得到的像应为（）。A.正立放大的虚像B.倒立缩小的实像C.倒立放大的实像D.正立等大的实像作答：________6.（3分）玻璃的折射率为1.50，光从玻璃射向空气时发生全反射的临界角满足（）。A.sinC=1.50B.sinC=1/1.50C.tanC=1.50D.cosC=1/1.50作答：________7.（3分）单缝衍射实验中，减小缝宽而保持单色光波长、屏距不变，中央亮纹的变化趋势是（）。A.变窄且更亮B.变宽且总能量更分散C.位置不变且宽度不变D.完全消失作答：________8.（3分）冬季从室外进入温暖房间，眼镜片上会出现小水珠。该现象主要涉及的物态变化和热量交换是（）。A.水蒸气液化并放热B.水蒸气汽化并吸热C.液态水凝固并放热D.冰升华并吸热作答：________9.（3分）一定质量理想气体在体积不变时温度从300K升到360K，压强变化为（）。A.变为原来的0.83倍B.变为原来的1.20倍C.保持不变D.变为原来的2倍作答：________10.（3分）质量相同的甲、乙两种液体吸收相同热量后，甲升温8℃，乙升温4℃。若无热损失，则（）。A.甲的比热容约为乙的2倍B.乙的比热容约为甲的2倍C.两者比热容相同D.无法比较比热容作答：________11.（3分）下列关于热传递方式的判断正确的是（）。A.真空保温层主要减弱热传导和对流B.黑色粗糙表面比银白光滑表面辐射能力弱C.电热水壶底部加热水主要依靠辐射传遍全壶D.金属勺柄变热主要是对流结果作答：________12.（3分）教室内用超声波测距仪测墙距，同时用激光指示器瞄准墙面。若室温升高而仪器未做声速补偿，则超声测得的距离通常会（）。A.偏小，因为实际声速变大B.偏大，因为实际声速变小C.不变，因为回波时间不变D.无法判断，因为光速变化显著作答：________二、填空题（本大题共6小题，每小题4分，共24分）13.（4分）声强级每增大10dB，声强约变为原来的______倍；若两个独立同频声源在某点声强相同，合成声强级比单个声源约增大______dB。________________________________________________________________________________________________________________________________________________________________________________14.（4分）在空气中向峭壁发出短促声，0.60s后听到回声，声速取340m/s，则人与峭壁距离为______m；若温度升高导致声速增大而仍用340m/s计算，所得距离将______（填“偏大”“偏小”或“不变”）。________________________________________________________________________________________________________________________________________________________________________________15.（4分）焦距为10cm的凸透镜，当物距为30cm时，像距为______cm；该像相对于物体是______（填“放大”“缩小”或“等大”）的实像。________________________________________________________________________________________________________________________________________________________________________________16.（4分）折射率为1.50的玻璃射向空气时，临界角C满足sinC=______；C约为______°（保留一位小数）。________________________________________________________________________________________________________________________________________________________________________________17.（4分）0.20kg水温度由18℃升至28℃，水的比热容取4.2×10³J/(kg·℃)，吸收热量为______J；若加热器实际消耗电能1.2×10⁴J，则加热效率约为______%。________________________________________________________________________________________________________________________________________________________________________________18.（4分）一定质量理想气体在压强不变时，温度由300K升至450K，体积由3.0L变为______L；若用摄氏温度直接成比例计算，会产生______（填“系统性”或“偶然性”）错误。________________________________________________________________________________________________________________________________________________________________________________三、实验与材料分析题（本大题共5小题，每小题6分，共30分）19.（6分）某小组用共鸣管测量空气中的声速。音叉频率为500Hz，第一次、第二次听到明显共鸣时管内空气柱长度分别为17.0cm、51.2cm。忽略管口修正差异，回答下列问题。（1）根据相邻共鸣长度差求声波波长。（2）计算空气中的声速，并说明结果与常温空气声速是否相符。（3）指出一项使测量值偏离真实值的实验因素。本题作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________20.（6分）在“测定凸透镜焦距并研究成像规律”的实验中，某透镜成像记录如下表。物距u/cm30.020.015.0像距v/cm15.020.030.0（1）任选一组数据估算透镜焦距。（2）当物距为20.0cm时，像的性质是什么？（3）若将物体从30.0cm慢慢移到15.0cm，像距和像的大小如何变化？本题作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________21.（6分）阅读材料：某保温杯内胆采用双层金属壁，中间抽成近似真空，内壁表面较光亮，杯盖处使用低导热材料。把0.50kg、90℃的热水倒入杯中，1h后降至82℃。水的比热容取4.2×10³J/(kg·℃)。（1）计算1h内水放出的热量。（2）分别说明真空层、光亮内壁和杯盖材料主要减弱哪一种热传递。（3）若只从杯口敞开处散热增强，水温下降会怎样变化？本题作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________22.（6分）光纤通信中，纤芯折射率略大于包层折射率，光在纤芯与包层交界面多次全反射而沿光纤传播。设纤芯折射率n₁=1.48，包层折射率n₂=1.46。（1）写出纤芯到包层界面的临界角表达式并计算临界角的正弦值。（2）说明为什么包层折射率必须略小于纤芯折射率。（3）若光纤弯折过急，通信质量可能下降，请从入射角角度解释。本题作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________23.（6分）某超声测距模块在室内测量桌面到天花板的高度。模块发出超声脉冲到接收回波的时间间隔为12.0ms。声速近似满足v=331+0.60t，其中t为摄氏温度。测量时室温为25℃。（1）求当时空气中的声速。（2）求桌面到天花板的高度。（3）若误按20℃的声速进行计算，结果相对正确值偏大还是偏小？偏差约为多少？本题作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________四、综合计算与应用题（本大题共5小题，每小题6分，共30分）24.（6分）一列救护车警报器发出频率为850Hz的声音，车以20m/s速度靠近静止观察者，空气中声速取340m/s。（1）观察者听到的频率是多少？（2）车辆远离观察者时，观察者听到的频率是多少？（3）解释同一警报器靠近和远离时音调不同的物理原因。本题作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________25.（6分）水池中有一点光源位于水面下1.20m处，水的折射率取4/3。只考虑水面到空气的折射与全反射。（1）求水到空气的临界角的正弦值。（2）估算光能从水面直接射出的圆形区域半径。可用tan48.6°≈1.13。（3）从空气中竖直向下看，该点光源的视深约为多少？本题作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________26.（6分）一定质量理想气体经历一个循环过程：每一循环从高温热源吸收600J热量，向低温热源放出420J热量。装置每秒完成8个循环。（1）求每一循环对外做功。（2）求热机效率。（3）若全部输出功用于竖直提升20kg物体，g取10N/kg，求1s内最多可提升的高度。本题作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________27.（6分）某简易太阳能加热装置用透镜会聚太阳光。透镜有效面积为0.20m²，太阳辐照度按800W/m²计，装置把入射能量的50%转化为水的内能。现加热0.50kg水，使其由20℃升至80℃，水的比热容取4.2×10³J/(kg·℃)。（1）求水需要吸收的热量。（2）求装置对水的有效加热功率。（3）估算加热所需时间，并指出实际时间常大于计算值的一项原因。本题作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________28.（6分）一套声光热综合检测装置用金属直尺作光路基准，直尺原长2.000m，线膨胀系数为1.2×10⁻⁵/℃。实验室温度从20℃升到45℃，同时利用声脉冲测量两端距离。空气声速按v=331+0.60t计算。（1）求金属直尺的伸长量。（2）求45℃时空气中的声速。（3）若声脉冲往返时间为11.6ms，按45℃声速计算得到的距离是多少？并说明声学距离与金属尺读数不完全一致时，应优先检查哪两类因素。本题作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________整卷计算与作图草稿区供声速换算、透镜成像、热量计算、单位整理与必要作图使用；正式得分以各题作答区书写内容为准。_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________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参考答案与解析说明：参考答案按题号逐题对应，理科计算题给出主要公式、代入、单位和结论；不同合理解法可按相同评分要点给分。一、单项选择题123456789101112AABABBBABBAA1.答案：A关键公式或判据：λ=v/f；机械波传播需要介质。解析：波长λ=v/f=343/500m，约为0.686m。频率由振源决定，音叉每秒振动500次；声波是机械波，不能在真空中传播；声强与振幅、传播距离等有关，不能只由频率决定。2.答案：A解析：音叉的振动频率由自身结构和受激方式决定，进入不同介质后频率保持不变；声速改变会使波长改变。水中声速通常大于空气中，所以波长更长；音调主要由频率决定，不因介质改变而必然降低。3.答案：B关键公式或判据：相邻共鸣长度差ΔL=λ/2；v=fλ。解析：相邻共振管长差等于半个波长，故λ/2=0.172m，λ=0.344m；声速v=fλ=1000×0.344m/s=344m/s。4.答案：A关键公式或判据：折射定律n₁sini=n₂sinr；跨介质频率不变。解析：光由空气进入玻璃时传播速度减小，折射角小于入射角，折射光线靠近法线；光的频率由光源决定，跨介质传播时频率不变，波长随速度改变。5.答案：B关键公式或判据：物距u>2f时，成倒立、缩小、实像。解析：由凸透镜成像规律，物距大于二倍焦距时，像位于一倍焦距到二倍焦距之间，为倒立、缩小、实像。本题30cm大于20cm。6.答案：B关键公式或判据：全反射临界角sinC=n₂/n₁。解析：光从折射率较大的玻璃射向空气才可能全反射，临界角满足sinC=n₂/n₁=1/1.50。7.答案：B解析：单缝衍射中央亮纹角宽度与波长成正比、与缝宽成反比；缝宽减小时中央亮纹变宽，同时能量在更宽区域分布，亮度会降低。8.答案：A解析：室内暖湿空气接触温度较低的镜片，水蒸气遇冷液化成小水珠，液化过程放出热量。9.答案：B关键公式或判据：等容过程p/T=常量。解析：体积不变的一定质量理想气体满足p/T为常量，温度由300K升至360K，压强变为原来的360/300=1.20倍。10.答案：B关键公式或判据：Q=cmΔT。解析：由Q=cmΔT，质量和吸热相同，则比热容与升温成反比。乙升温为甲的一半，说明乙的比热容约为甲的2倍。11.答案：A解析：真空层没有普通物质，主要抑制热传导和对流；银白光滑表面辐射能力弱，黑色粗糙表面辐射能力强；水壶内热量主要靠对流分布，金属勺柄变热主要靠热传导。12.答案：A解析：室温升高时实际声速增大。仪器若仍按较小声速换算，利用s=vt/2得到的距离小于真实距离，因此读数偏小。二、填空题13.答案：10；约3关键公式或判据：L=10lg(I/I₀)。解析：声强级L=10lg(I/I₀)。声强增大10倍，声强级增大10dB；两个声强相同且独立叠加时总声强为2I，增量10lg2≈3dB。14.答案：102；偏小关键公式或判据：回声测距s=vt/2。解析：回声往返路程为2s，s=vt/2=340×0.60/2=102m。若真实声速比340m/s大而仍取340m/s，计算路程偏小，距离也偏小。15.答案：15；缩小关键公式或判据：薄透镜公式1/f=1/u+1/v。解析：薄透镜公式1/f=1/u+1/v，取f=10cm，u=30cm，得1/v=1/10-1/30=1/15，所以v=15cm。像距小于物距，成倒立缩小实像。16.答案：2/3；41.8关键公式或判据：sinC=n₂/n₁。解析：临界角满足sinC=n₂/n₁=1/1.50=2/3，对应C≈41.8°。17.答案：8.4×10³；70关键公式或判据：Q=cmΔT，η=Q有效/E输入。解析：Q=cmΔT=4.2×10³×0.20×10J=8.4×10³