题目内容
| ××滚筒式洗衣机 | |
| 电压/V | 220 |
| 洗衣功率/W | 200 |
| 脱水功率/W | 330 |
| 最大洗涤容量/kg | 5 |
| 水加热功率/W | 2000 |
（1）求脱水时的正常工作电流．
（2）某次洗衣时，首次进水量为15L，正常加热15min水温达到设定要求．
①求首次进水的质量；
②求加热过程消耗的电能；
③若加热时电能的84%转化为水的内能，求水升高的温度．
【答案】分析：（1）已知洗衣机的脱水功率和额定电压，根据I=
即可求出脱水时的正常工作电流；
（2）①已知首次进水的体积，根据m=ρv即可求出首次进水的质量；
②已知加热时的功率和加热时间，根据W=Pt即可求出加热过程消耗的电能；
③先求出水吸收的热量，然后根据Q=cm△t的变形公式△t=
即可求出水升高的温度．
解答：解：（1）已知脱水时的功率P=330W，电压U=220V，
∴脱水时的正常工作电流：I=
=
=1.5A
答：洗衣机脱水时的正常工作电流是1.5A．
（2）①m水=ρ水V水=1.0×103kg/m3×1.5×10-2m3=15kg；
②加热时的功率为2000W，加热时间为15min，
∴加热过程消耗的电能
W=Pt=2000W×15×60s=1.8×106J；
③加热时电能的84%转化为水的内能
∴Q水吸=W电×84%=1.8×106J×84%=1.512×106J
∴水升高的温度
△t=
=
=24℃
答；①首次进水的质量为15kg；
②加热过程消耗的电能为1.8×106J；
③水升高的温度为24℃．
点评：（1）本题综合考查了密度公式的应用、热量计算公式的应用以及电功的计算公式的应用情况，是一道综合性较强的计算题．
（2）本题中学生在计算水的质量时容易出错，一定要记住：1L=1dm3．
即可求出脱水时的正常工作电流；（2）①已知首次进水的体积，根据m=ρv即可求出首次进水的质量；
②已知加热时的功率和加热时间，根据W=Pt即可求出加热过程消耗的电能；
③先求出水吸收的热量，然后根据Q=cm△t的变形公式△t=
即可求出水升高的温度．解答：解：（1）已知脱水时的功率P=330W，电压U=220V，
∴脱水时的正常工作电流：I=
==1.5A答：洗衣机脱水时的正常工作电流是1.5A．
（2）①m水=ρ水V水=1.0×103kg/m3×1.5×10-2m3=15kg；
②加热时的功率为2000W，加热时间为15min，
∴加热过程消耗的电能
W=Pt=2000W×15×60s=1.8×106J；
③加热时电能的84%转化为水的内能
∴Q水吸=W电×84%=1.8×106J×84%=1.512×106J
∴水升高的温度
△t=
==24℃答；①首次进水的质量为15kg；
②加热过程消耗的电能为1.8×106J；
③水升高的温度为24℃．
点评：（1）本题综合考查了密度公式的应用、热量计算公式的应用以及电功的计算公式的应用情况，是一道综合性较强的计算题．
（2）本题中学生在计算水的质量时容易出错，一定要记住：1L=1dm3．
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