1、19.1 比熱(Specific Heat)mmT=1OCQ=?19.1 比熱(Specific Heat)mmT=1OCQ=?19.1 比熱(Specific Heat)cmTQ=?mc19.1 比熱(Specific Heat)T1TfT2m1:c1 :T1=T1-Tfm2=c2=T2=Tf -T2Q1=m1 c1(T1-Tf)Q2=m2 c2(Tf T2)Q1=Q2m1 c1(T1-Tf)=m2 c2(Tf T2)TT(m)TT(cmcf112f221 Sol:m1c1(T1-Tf)=m2c2(Tf-T2)+m3 c3(Tf-T2)0.08 450(200-Tf)=0.25 4190 (
2、Tf-20)+0.1 390 (Tf-20)Tf =25.8 C19.2 潛熱(Latent Heat)19.2 潛熱(Latent Heat)19.2 潛熱(Latent Heat)相變時：物質的溫度與熱量變化(潛熱)SOL:觀念：觀念：2kg的冰所吸收的熱量的冰所吸收的熱量 5kg的溫水所放出的熱量的溫水所放出的熱量 m1c1T1+m1L1+m1Cw(Tf-0)=mw cw(45-Tf)(-100 00)(融化融化)(00 Tf)(450 Tf )2 2100 (10)+2 3.34 105+2 4200 (Tf)=54200(45-Tf)Tf =80 C(已知冰的溶解熱:L3.34105 J/kg)1cal=4.2 Joul(1)重物下移重物下移(h):損失位能損失位能=重力對物體所作的功重力對物體所作的功 W=mgh(2)水溫變化水溫變化(T):溫度上升時，所吸收的熱量溫度上升時，所吸收的熱量 Q=mcT(3)熱與功的關係式熱與功的關係式:)kcal/joul(4200)cal/joul(2.4K)cal/joul(2.4TmchmgQW 熱功當量：熱功當量：(a)(b)(a)(b)(a)(b)(a)(b)AdQ/dt LTHTCR1R2T2T1A T RR1R2 R RR1R2T2T1A T R
