(1)∵ρ=,V=125L=0.125m3, ∴m=ρV=1.0×103kg/m3×0.125m3=125kg; (2)Q吸=5.25×107J×40%=2.1×107J, ∵Q吸=c水m△t, ∴水吸热后温度升高值: △t==2.1×107J | 4.2×103J/(kg•℃)×125kg | =40℃; (3)同样质量、初温的水加热到同样的温度需的时间: ∵P=, ∴t===35000s=9h40min20s. 答;(1)该热水器最多能装125千克的水; (2)那么可以使水的温度上升40摄氏度; (3)需要9h40min20s. |