Răspuns:
[tex]\text{Cand un corp lichid sau solid se incalzeste/raceste, formula pentru caldura primita/cedata este:}\\\Delta Q \text{ (sau $Q$)}=mc\Delta t, \text{ unde}\\\\m=\text{masa corpului studiat}\\c=\text{caldura specifica a corpului studiat}\\\Delta t=\text{variatia/diferenta de temperatura}[/tex]
[tex]\text{Cand un corp isi schimba starea de agregare, adica se topeste/solidifica sau se }\\\text{vaporizeaza/condenseaza, caldura primita/cedata are formula:}\\\Delta Q \text{ (sau $Q$)}=m\lambda, \text{ unde}\\\\m=\text{masa partii din corpul studiat care sufera transformarea}\\\lambda=\text{caldura latenta de topire (solidificare) sau de vaporizare (condensare)}[/tex]
[tex]\text{Caldura cedata este negativa (deoarece variatia/diferenta/gradientul de temperatura este}\\\text{negativ, variatia de energie este negativa), iar caldura primita este pozitiva (deoarece }\\\text{variatia/diferenta/gradientul de temperatura este pozitiv, variatia de energie este pozitiva).}[/tex]
[tex]\text{\bf Problema 1}\\\text{Date: }\\m=100\,g;\:t_1=10\,^{\circ}C; \:t_2=60\,^{\circ}C; c=4200\,J/(kg\cdot K)\\\\\text{Rezolvare:}\\\Delta t=t_2-t_1=60\,^{\circ}C-10\,^{\circ}C=50\,^{\circ}C=50\,K\\\text{Se incalzeste un corp lichid}\implies\Q_{primit}=mc\Delta t\\Q_{primit}=0,1\,kg\cdot 4200 \,J/(kg\cdot K)\cdot 50\,K=21000 \,J[/tex]
[tex]\text{\bf Problema 2}\\\text{Date:}\\Q=170\,kJ;\: \lambda=850\,kJ/kg\\\\\text{Rezolvare:}\\\text{O masa dintr-un corp lichid isi schimba starea de agregare (se vaporizeaza)}\implies\Q=m\lambda \implies m=\frac{Q}{\lambda}\\m=\frac{170\,kJ}{850\,kJ/kg}=0,2\,kg[/tex]
[tex]\text{\bf Problema 3}\\\text{Date:}\\m=500\,g;\:t_0=-10\,^{\circ}C;\:t_1=0\,^{\circ}C\text{ (temperatura la care se topeste gheata)};\\t_2=20\,^{\circ}C;\:c_{gheata}=2100\,J/(kg\cdot K);\:\lambda=333\,kJ/kg;\:c_{apa}=4200\,J/(kg\cdot K)\\\\\text{Rezolvare:}\\\text{b) Primul proces este cel de incalzire al ghetii (corp solid): }\\\Delta t_1=t_1-t_0=0\,^{\circ}C-(-10\,^{\circ}C)=10\,^{\circ}C=10\,K\\Q_1=mc_{gheata}\Delta t_1\\Q_1=0,5\,kg\cdot 2100\,J/(kg\cdot K)\cdot 10\,K=10500\, J=10,5\,kJ\\[/tex]
[tex]\text{Al doilea proces este cel de topire (schimbarea starii de agregare) a ghetii:}\\Q_2=m\lambda\\Q_2=0,5\,kg\cdot 333\,kJ/kg=166,5\,kJ\\[/tex]
[tex]\text{Al treilea proces este cel de incalzirea a apei (corp lichid) rezultata din topirea ghetii:}\\\Delta t_2=t_2-t_1=20\,^{\circ}C-0\,^{\circ}C=20\,^{\circ}C=10\,K\\Q_3=mc_{apa}\Delta t_2\\Q_3=0,5\,kg\cdot 4200\,J/(kg\cdot K)\cdot 20\,K=42000\, J=42\,kJ\\\\Q_{consumat}=Q_1+Q_2+Q_3\\Q_{consumat}=10,5\,kJ+166,5\,kJ+42\,kJ=219\,kJ[/tex]
