质量为500g的铝块,温度从20℃升高到100℃,共吸收多少热量?[C铝=0.88×10^3J/(kg×℃)
人气:256 ℃ 时间:2019-10-03 22:57:32
解答
Q=cm(t-t0)=0.88*103*0.5*(100-20)=3.52*104J
我没写单位了,自己添上
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