消耗的电能：=220r ÷6400r/(kWh) = 220/6400 kWh = 0.034375 kWh = 123.75kJ
水的比热容是 4.2kJ/(kg℃)
Q=△m × Cp ×△T
△m = Q÷Cp÷△T
= 123.75 kJ ÷ 4.2kJ/(kg℃)÷ 10℃
= (123.75÷42) kg
= 2.946 kg
一个家用电能表的盘面上标有“6400 r／kW·h”字样,若某个电热水器单独使用时,电表转盘在100s内转了220转,则热水器消耗的电能是多少?这个热水器放出的热量能使多少质量的水温度升高10℃?[ 假设没有能量损失]
消耗的电能：=220r ÷6400r/(kWh) = 220/6400 kWh = 0.034375 kWh = 123.75kJ
水的比热容是 4.2kJ/(kg℃)
Q=△m × Cp ×△T
△m = Q÷Cp÷△T
= 123.75 kJ ÷ 4.2kJ/(kg℃)÷ 10℃
= (123.75÷42) kg
= 2.946 kg