Question

3£®ÈçͼËùʾΪÍøÉÏÈÈÏúµÄÒ»¿î¹ú²úµçÈÈË®ºø£¬ÎªÁËÑéÖ¤Æä¹ã¸æËùÊöÈÈЧÂʵÄÕæʵÐÔ£¬Ð¡ºìͬѧ¾ö¶¨½øÐÐʵÑé̽¾¿£®±íÒ»ÊǵçÈÈË®ºøµÄÃúÅÆ£º±í¶þÊÇСºì̽¾¿¸ÃµçÈÈË®ºøЧÂÊËù¼Ç¼µÄÊý¾Ý£ºÒÑ֪ˮµÄ±ÈÈÈÈÝΪc_Ë®=4.2¡Ál0³J/£¨kg•¡æ£©£®Çó£º
±íÒ»£º

²úÆ·ÐͺÅ	SWF12P1A-150
¶î¶¨µçѹU/V	220
µçԴƵÂÊf/Hz	50
¶î¶¨¹¦ÂÊP/W	1200
ÈÝÁ¿V/L	1.2

±í¶þ£º

µçÔ´µçѹUµç/V	220
ºøÄÚË®µÄÖÊÁ¿m/kg	1
Ë®µÄ³õÎÂt1/¡æ	33
Ë®µÄ·Ðµãt2/¡æ	98
Ë®´Ó³õμÓÈÈÖÁ¸Õ·ÐÌÚµÄʱ¼ät/s	250

£¨1£©µçÈÈË®ºøµÄµç×裨±£Áô1λСÊý£©£»
£¨2£©µ±µçÈÈË®ºøʵ¼Ê¹¤×÷µçѹΪ110Vʱ£¬ËüµÄʵ¼Ê¹¦ÂÊ£¨±£ÁôÕûÊý£©£»
£¨3£©µçÈÈË®ºøµÄЧÂÊ£®

Answer 1

·ÖÎö £¨1£©ÓÉÃúÅÆ¿ÉÖªµçÈÈË®ºøµÄ¹¤×÷µçѹºÍ¹¦ÂÊ£¬¸ù¾ÝP=$\frac{{U}^{2}}{R}$Çó³öµçÈÈË®ºøµÄµç×裻
£¨2£©ÖªµÀµçÈÈË®ºøµÄµç×èºÍʵ¼Êµçѹ£¬¸ù¾ÝP=$\frac{{U}^{2}}{R}$Çó³öµçÈÈË®ºøµÄʵ¼Ê¹¦ÂÊ£»
£¨3£©¸ù¾ÝQ_Îü=cm£¨t-t₀£©Çó³öË®ÎüÊÕµÄÈÈÁ¿£¬¸ù¾ÝW=PtÇó³öÏûºÄµÄµçÄÜ£¬ÀûÓÃЧÂʹ«Ê½Çó³öµçÈÈË®ºøµÄЧÂÊ£®

½â´ð ½â£º£¨1£©ÓÉP=$\frac{{U}^{2}}{R}$¿ÉµÃ£¬µçÈÈË®ºøµÄµç×裺
R=$\frac{{U}^{2}}{P}$=$\frac{£¨220V£©^{2}}{1200W}$¡Ö40.3¦¸£»
£¨2£©µ±µçÈÈË®ºøʵ¼Ê¹¤×÷µçѹΪ110Vʱ£¬ËüµÄʵ¼Ê¹¦ÂÊ£º
P_ʵ=$\frac{{U}_{ʵ}^{2}}{R}$=$\frac{£¨110V£©^{2}}{\frac{121}{3}¦¸}$=300W£»
£¨3£©Ë®ÎüÊÕµÄÈÈÁ¿£ºQ_Îü=c_Ë®m£¨t-t₀£©=4.2¡Á10³J/£¨kg•¡æ£©¡Á1kg¡Á£¨98¡æ-33¡æ£©=2.73¡Á10⁵J£¬
µçË®ºø250sÏûºÄµÄµçÄÜ£º
W=Pt=1200W¡Á250s=3¡Á10⁵J£¬
¼ÓÈÈЧÂÊ£º
¦Ç=$\frac{{Q}_{Îü}}{W}$¡Á100%=$\frac{2.73¡Á1{0}^{5}J}{3¡Á1{0}^{5}J}$¡Á100%=91%£®
´ð£º£¨1£©µçÈÈË®ºøµÄµç×èԼΪ40.3¦¸£»
£¨2£©µ±µçÈÈË®ºøʵ¼Ê¹¤×÷µçѹΪ110Vʱ£¬ËüµÄʵ¼Ê¹¦ÂÊΪ300W£»
£¨3£©µçÈÈË®ºøµÄЧÂÊΪ91%£®

µãÆÀ ±¾Ì⿼²éÁ˵繦Âʹ«Ê½µÄÁé»îÓ¦ÓúÍÎüÈȹ«Ê½¡¢µç¹¦¹«Ê½¡¢Ð§Âʹ«Ê½µÄÓ¦Ó㬹ؼüÊÇ´Ó±í¸ñÖеóöÏà¹ØµÄÐÅÏ¢£®