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图中是某太阳能热水器,向其中注入50kg的水,阳光照射﹣段时间后,水温从10℃升高到50℃.水的比热容是4.2×103J/(kg•℃).试求:

(1)这段时间该热水器中的水吸收的热量是多少?

(2)如果这段时间该太阳能热水器接收到太阳辐射的热量是2.8×107J,则这段时间该热水器的效率是多少?

(3)如果水吸收的热量用天然气来提供,需要完全燃烧多少m3的天然气;(天然气的热值为8.4×l07J/m3,天然气完全燃烧放出的热量全部给水吸收)

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答案(点此获取答案解析)

解:(1)水吸收的热量:

Q=cm(t﹣t0

=4.2×103J/(kg•℃)×50kg×(50℃﹣10℃)

=8.4×106J.

(2)∵热水器接收到太阳辐射的热量Q=2.8×107J,

∴这段时间该热水器的效率:

η=

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