Question

Calculati caldura specifica pentru vaporizarea a 300 grame de apa aflate la temperatura de 20 grade celsius

Answer 1

[tex]\displaystyle \text{Se da:}\\ \\ m=300g=0,3kg\\ \\ t_1=20^\circ\\ \\ c=4190\frac{J}{kg\times K}\\ \\ t_f=100^\circ\\ \\ \lambda_v=2260\frac{kJ}{kg}=2260000\frac{J}{kg}\\ \\ Q=?J\\ \\ \\[/tex]


[tex]\text{Formule:}\\ \\ \text{Pentru a evapora apa ea trece 2 stari:}\\ \\ \text{1)Aducerea ei la temperatura de fierbere (}t_f\text{)}\\ \\ \text{2)Evaporarea propriu-zisa}\\ \\ \text{Deci impartim caldura necesara in 2:}\\ \\ \\ Q=Q_1+Q_2\\ \\ \\[/tex]


[tex]Q_1=c\times m\times\Delta t\\ \\ Q_1=c\times m\times(t_f-t_1)\\ \\ \\ Q_2=m\times\lambda_v\\ \\ \\ Q=c\times m\times(t_f-t_1)+m\times\lambda_v\\ \\ Q=m\times[c\times(t_f-t_1)+\lambda_v]\\ \\ \\ \text{Calcule:}\\ \\ Q=0,3\times[4190\times(100-20)+2260000]=778560J[/tex]