(1)∵ρ=[m/V],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=
Q吸
c水m=
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=[W/t],
∴t=
Q吸
P=
5.25×107J
1500W=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.