10吨80摄氏度度的热水,加入多少吨34摄氏度的温水,可以使水温最终达到45摄氏度

最好能给出计算公式,并能进行详细的解释,急用
2025-03-26 09:14:12
推荐回答(5个)
回答1:

80度的热水,加入34度的温水,使水温达到45度.相当于热水放热,冷水吸热,放热量=吸热量.
水的比热为C
Q(放)=Q(吸)
C*m(热水)*(80-45)=C*m(冷水)*(45-34)
m(冷水)=(35/11)*m(热水)=350/11(吨)=31.82(吨)

回答2:

设加入34度的水X吨,混合以后,水温为45度。
列式:
((10×80)+(34×X))/(10+X)=45
计算出X=350/11=31.82吨

分子里,10×80和34×X可以理解为两种水对混合后水温的贡献.
分母10+X,是混合后水的重量.

回答3:

34度到80度水的比热变化不大,可以认为相等
45-34=11
80-45=35
根据热量衡算,吸热等于放热,有
10*35/11=31.81818吨

回答4:

设要用水为x
10*80*C+x*34*C=(10+x)*45*C
解得 x=350/11吨

回答5:

没有说明加几次水 分批次加水 可以提高效率

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