2026年浙教版适配专升本物理暑假衔接卷声光热综合应用标准试卷第322套（含答案解析与可打印作答区）

专升本物理暑假衔接卷声光热综合应用2026年浙教版适配专升本物理暑假衔接卷声光热综合应用标准试卷第322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分适用对象专升本暑假衔接内容范围声学、光学、热学综合应用注意事项1.本卷共28题，满分120分，考试时间120分钟。所有答案写在相应作答区内。2.单项选择题每题只有一个最佳答案；填空题要求写出准确数值、单位或关键词。3.计算题应写出主要公式、代入过程、单位和结论；只写最终结果不得满分。4.作答时保持卷面整洁，严禁抄袭、夹带或使用不符合规定的资料。试题结构与分值题型题号题量每题分值小计单项选择题1—1212题3分36分填空题13—186题4分24分实验与材料题19—235题6分30分综合应用题24—285题6分30分合计1—2828题120分一、单项选择题（本题共12小题，每小题3分，共36分）请将正确选项填入下方答题栏。题号123456789101112答案1.（3分）某弦乐器一根琴弦两端固定，弦长0.60m，调到基频440Hz。不计弦的伸长和空气阻力，琴弦上横波的传播速度最接近下列哪一项？A.264m/sB.440m/sC.528m/sD.880m/s2.（3分）在同一位置先开启一只同型号扬声器，测得声强级为70dB；再同时开启另一只与它相同且相互独立的扬声器。若两只扬声器到测点距离相同且相位关系不固定，声强级大约变为A.70dBB.73dBC.76dBD.140dB3.（3分）一列频率为1000Hz的声波在空气中传播，声速取340m/s。观察者以17m/s速度迎着静止声源运动，观察者听到的频率约为A.950HzB.1000HzC.1050HzD.1100Hz4.（3分）一束光从空气斜射入平行玻璃砖后再从另一侧射出。若不计反射损失，下列关于出射光的说法正确的是A.出射光与入射光平行B.出射光一定沿法线方向C.出射角一定小于入射角D.出射光频率变小5.（3分）焦距为10cm的薄凸透镜前方30cm处放一小物体，在另一侧得到清晰像。该像的位置和性质为A.距透镜15cm，倒立缩小实像B.距透镜15cm，正立放大虚像C.距透镜30cm，倒立等大实像D.距透镜20cm，倒立放大实像6.（3分）折射率为1.50的玻璃中有一束光射向玻璃与空气的界面。若要发生全反射，入射角应满足的最小临界条件约为A.大于24.6°B.大于33.7°C.大于41.8°D.大于48.6°7.（3分）把0.20kg、80℃的水与0.30kg、20℃的水充分混合，不计容器吸热和热量散失，混合后水温约为A.36℃B.40℃C.44℃D.50℃8.（3分）一定质量理想气体在温度不变时体积变为原来的一半。下列说法正确的是A.压强变为原来的一半B.压强变为原来的2倍C.内能增加一倍D.气体对外做功为正9.（3分）一根2.0m长的钢尺温度升高50℃，钢的线膨胀系数取12×10⁻⁶/℃。钢尺长度增加约为A.0.12mmB.0.60mmC.1.2mmD.12mm10.（3分）两块材料、面积和温差都相同的平板隔热层，厚度分别为d和2d。在稳定传热状态下，厚度为2d的隔热层单位时间传热量约为厚度为d的A.1/4B.1/2C.2倍D.4倍11.（3分）光在折射率为1.50的光纤中传播10km，真空光速取3.0×10⁸m/s。仅考虑传播时间，所需时间约为A.5.0μsB.50μsC.0.50msD.5.0ms12.（3分）某超声测距仪在20℃空气中向墙面发射脉冲，接收回波的总时间为0.120s。声速按v=331+0.6t（m/s）估算，则墙面到测距仪的距离最接近A.10.3mB.20.6mC.34.3mD.41.2m二、填空题（本题共6小题，每小题4分，共24分）13.（4分）频率为500Hz的声波在空气中的传播速度取340m/s，则波长为______m；若声波进入传播速度更大的介质而频率不变，则波长将______。答：________________________________________________14.（4分）一片薄凸透镜的焦距为25cm，其光焦度为______D。若用它会聚平行光，焦点应位于透镜后方______cm处。答：________________________________________________15.（4分）光在某透明介质中的速度为2.0×10⁸m/s，真空光速取3.0×10⁸m/s，则该介质折射率为______；光由空气进入该介质时频率______。答：________________________________________________16.（4分）理想气体吸收热量Q，对外做功W，内能变化量为______。在只发生等温变化的理想气体过程中，内能变化量为______。答：________________________________________________17.（4分）用电热法测液体比热容时，电压U、电流I、加热时间t、液体质量m、温度升高量ΔT均已测得，不计热损失时比热容表达式为______；实验中可通过______来减小热量散失。答：________________________________________________18.（4分）物体的热辐射强弱与温度密切相关。温度升高时，单位面积单位时间向外辐射的能量通常______；黑色粗糙表面比银白光滑表面对热辐射的吸收能力______。答：________________________________________________填空题草稿与单位核对区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________三、实验与材料题（本题共5小题，每小题6分，共30分）19.（6分）某学习小组用共鸣管测定空气中的声速。音叉频率为500Hz，调节水面位置时，第一次和第二次出现明显共鸣的空气柱长度分别为17.0cm和51.0cm。（1）写出利用相邻两次共鸣长度差求波长的关系式。（2）计算空气中的声速。（3）说明用两次共鸣长度差计算能削弱哪一类系统误差。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________20.（6分）在探究凸透镜成像规律时，某透镜固定在光具座40.0cm刻度处。移动蜡烛和光屏得到下表数据。实验序号物距u/cm像距v/cm像的性质130.015.0倒立、缩小、实像220.020.0倒立、等大、实像315.030.0倒立、放大、实像（1）由第2组数据判断透镜焦距。（2）用第1组数据验证薄透镜公式。（3）若物体从30.0cm处逐渐移到15.0cm处，像距和像的大小如何变化？作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________21.（6分）用量热法估测某金属的比热容。将0.40kg金属块在沸水中充分加热后迅速放入0.20kg、20℃的水中，最终水温为28℃。已知水的比热容为4.2×10³J/(kg·℃)，不计容器吸热。（1）写出热平衡方程。（2）计算该金属的比热容。（3）若金属转移过程中已向空气散热，计算结果与真实值相比偏大还是偏小？说明理由。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________22.（6分）阅读材料：城市隧道维护中常同时使用声学、光学和热学方法。工作人员用短促声脉冲判断障碍位置，用激光准直检查轨道旁标尺是否偏移，用红外测温仪寻找电缆接头异常发热点。某次检测时，声脉冲从检测车发出到收到障碍物回波共0.080s，当时空气温度为25℃；激光经过一块平行防护玻璃后照到标尺上；红外测温发现某接头温度高于周围20℃。声速仍按v=331+0.6t估算。（1）障碍物离检测车约多远？（2）平行防护玻璃会不会改变出射激光的传播方向？会产生什么可观察影响？（3）从热学角度说明电缆接头发热异常可能意味着什么。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________23.（6分）某实验台同时配有超声温度补偿和红外测温。超声探头正对反射板，探头到板的距离为1.715m。不同温度下测得往返时间如下表。空气温度/℃02040往返时间/ms10.3610.009.66（1）用20℃数据计算声速。（2）判断温度升高时声速如何变化。（3）若用红外测温仪测量亮银色金属表面，读数常偏低，提出一种改进方法。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________四、综合应用题（本题共5小题，每小题6分，共30分）24.（6分）一台倒车超声雷达在15℃环境中测量墙面距离。雷达发出脉冲到接收回波的时间为0.064s。若系统误按20℃声速进行计算，比较测得距离与真实距离。声速按v=331+0.6t（m/s）取值。（1）求15℃时的真实距离。（2）求系统按20℃声速得到的距离。（3）求距离误差，并判断偏大还是偏小。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________25.（6分）一台简易成像仪用焦距50mm的薄凸透镜给高80mm的标尺成像。标尺到镜头距离为1.00m。传感器可沿光轴微调。（1）求清晰成像时像距。（2）求像的高度。（3）若传感器固定在50mm处而不微调，为什么远处物体能较清晰而1.00m处标尺会变模糊？作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________26.（6分）某太阳能热水器集热面积为2.0m²。晴天太阳辐照度平均为750W/m²，连续照射3.0h后，水箱中50kg水由20℃升到55℃。水的比热容取4.2×10³J/(kg·℃)。（1）求水吸收的热量。（2）求太阳辐射输入到集热面的能量，并计算集热效率。（3）说明集热板做成黑色、外层加透明盖板的物理原因。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________27.（6分）两只相同扬声器S₁、S₂由同一信号源驱动，频率500Hz，相位相同，间距0.68m。空气中声速取340m/s。某点P到S₁、S₂的路程差记为Δr。（1）求声波波长。（2）当Δr=0.34m时，P点声强相对较大还是较小？说明理由。（3）若其中一只扬声器频率调为502Hz，听者在两声都较明显处会听到什么现象？频率是多少？作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________28.（6分）声光热一体化演示平台中，一根长2.0m的透明管两端开口，管内空气可被缓慢加热。用激光从空气以45°入射角射入折射率1.50的玻璃窗口；用扬声器激发管内基频共鸣。空气中声速按v=331+0.6t（m/s）估算。（1）计算20℃时两端开口管的基频。（2）计算空气温度升至45℃时基频的变化量。（3）求激光进入玻璃后的折射角，并说明光的频率是否改变。作答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

参考答案与解析本部分按题号逐题对应，供阅卷和订正使用。一、单项选择题123456789101112CBCAACCBCBBB1.答案：C。解析：两端固定弦的基频对应半个波长等于弦长，即λ=2L=1.20m。由v=λf得v=1.20×440=528m/s。A、B、D都没有同时满足基频与两端固定边界条件。提示：先判定固定弦基频，λ=2L。2.答案：B。解析：两个相同且相互独立的声源叠加时平均声强加倍，声强级增加10lg2≈3dB，因此70dB约变为73dB。不能把分贝数直接相加。提示：分贝为对数单位，声强加倍约增3dB。3.答案：C。解析：观察者迎向静止声源，接收频率f'=f(v+v_o)/v=1000×(340+17)/340≈1050Hz。A对应远离声源，B不计相对运动，D数值偏大。提示：观察者迎向声源，接收频率升高。4.答案：A。解析：在两表面平行的玻璃砖中，第一次折射向法线偏折，第二次折射背离法线偏折，两次偏折角相互抵消，出射光与入射光平行但发生侧向位移；光频率由光源决定，不因介质改变。提示：平行玻璃砖使光线平行出射并产生侧移。5.答案：A。解析：薄透镜成像满足1/f=1/u+1/v。f=10cm，u=30cm，解得v=15cm。物距大于二倍焦距，像为倒立缩小实像。提示：物距大于2f时，像距在f到2f之间。6.答案：C。解析：由玻璃射向空气时发生全反射的临界角满足sinC=n_2/n_1=1/1.50，C≈41.8°。入射角大于临界角时发生全反射。提示：全反射需从光密介质射向光疏介质。7.答案：C。解析：不计热损失，热水放热等于冷水吸热。平衡温度T=(0.20×80+0.30×20)/(0.50)=44℃。提示：混合水温按质量和比热加权。8.答案：B。解析：等温过程温度不变，理想气体内能不变。由pV恒定可知体积变为一半时压强变为2倍；气体被压缩时对外做功为负。提示：理想气体等温时内能不变，p与V成反比。9.答案：C。解析：线膨胀量ΔL=αLΔT=12×10⁻⁶×2.0×50m=1.2×10⁻³m=1.2mm。提示：线膨胀计算后注意单位换算。10.答案：B。解析：稳定导热的热流率与厚度成反比，厚度由d变为2d时，单位时间传热量变为原来的一半。提示：稳定导热热流率与厚度成反比。11.答案：B。解析：光纤中光速v=c/n=2.0×10⁸m/s，传播时间t=L/v=10000/(2.0×10⁸)=5.0×10⁻⁵s=50μs。提示：介质中光速应取c/n。12.答案：B。解析：20℃时声速v=331+0.6×20=343m/s。回波时间包含往返路程，所以s=vt/2=343×0.120/2≈20.6m。提示：回波时间对应往返路程，求距离要除以2。二、填空题13.答案：0.68；增大。解析：λ=v/f=340/500=0.68m。频率由声源决定，介质改变时频率不变；传播速度增大，波长随之增大。提示：介质改变不改变声源频率。14.答案：4；25。解析：光焦度P=1/f，其中f用米作单位。f=0.25m，P=4D；会聚平行光的焦点位于透镜后焦平面，距透镜25cm。提示：光焦度计算中焦距用米作单位。15.答案：1.5；不变。解析：折射率n=c/v=3.0×10⁸/(2.0×10⁸)=1.5。光进入介质后波速和波长改变，但频率由光源决定而不变。提示：光入介质后频率不变。16.答案：ΔU=Q-W；0。解析：按热力学第一定律，气体吸热Q、对外做功W时，内能变化ΔU=Q-W。理想气体内能只与温度有关，等温过程温度不变，所以ΔU=0。提示：本卷采用气体对外做功为正的符号约定。17.答案：c=UIt/(mΔT)；保温、加盖、缩短读数时间或搅拌后迅速读数。解析：电功UIt转化为液体吸收的热量mcΔT，故c=UIt/(mΔT)。保温和减少加热后的暴露时间可减小系统与环境的热交换。提示：电热法要先确认电功与吸热量的对应关系。18.答案：增强；更强。解析：温度越高，热辐射越强。黑色粗糙表面对辐射的吸收和发射能力通常强于银白光滑表面。提示：黑色粗糙表面通常吸收和发射辐射能力较强。三、实验与材料题19.参考答案与解析：（1）相邻两次共鸣的空气柱长度差等于半个波长，λ=2(L₂-L₁)。

（2）L₂-L₁=51.0cm-17.0cm=34.0cm=0.340m，λ=0.680m，v=fλ=500×0.680=340m/s。

（3）管口端部修正对两次长度读数几乎相同，取长度差后可大幅抵消端部修正带来的系统误差。评分要点：能写出相邻共鸣长度差与半波长的关系给2分；单位换算和声速计算给2分；说明端部修正抵消给2分。20.参考答案与解析：（1）第2组物距等于像距且成等大实像，此时u=v=2f=20.0cm，所以f=10.0cm。

（2）第1组中1/u+1/v=1/30.0+1