完全燃烧0.21kg的酒精所放出的热量,全部被质量为50kg、温度为20℃的

完全燃烧0.21kg的酒精所放出的热量,全部被质量为50kg、温度为20的水吸收,则水温升高  [q酒精=3.0×107J/kgc=4.2×103J/kg•℃]

答案

完全燃烧0.21kg的酒精放出的热量:

Q=qm酒精=3×107J/kg×0.21kg=6.3×106J

Q=cmt=Q

水温度升高值:

t===30

故答案为:30

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