finished section 5

main.tex

@@ -140,14 +140,22 @@ $L_{f}= \displaystyle{\frac{m_{\rm{H2O}} \cdot c_{\rm{H_2o}} \cdot \Delta\varthe
 
 $L_{f_1}=\displaystyle{\frac{\left(\left(m_1-m_{Kal}\right) \cdot 4182\frac{J}{kg \cdot K}\cdot \left(\vartheta_{T_1}-\vartheta_{T_2}\right)\right)-\left(\left(m_2-m_{1}\right)\cdot 4182 \frac{J}{kg \cdot K} \cdot \vartheta_{T_2}\right)}{m_2-m_{Kal}}} $\\\\
 
-$L_{1_{max}} =\displaystyle{\frac{\left(0.174\;kg \cdot 4182\frac{J}{kg \cdot K}\cdot 7.70^{\circ}C \right)-\left(0.184\;kg\cdot 4182 \frac{J}{kg \cdot K} \cdot 62.65^{\circ}C\right)}{0.184\;kg}} = 298302\frac{J}{kg}$\\\\
+$L_{f_{1_{max}}} =\displaystyle{\frac{\left(0.174\;kg \cdot 4182\frac{J}{kg \cdot K}\cdot 7.70^{\circ}C \right)-\left(0.010\;kg\cdot 4182 \frac{J}{kg \cdot K} \cdot 62.65^{\circ}C\right)}{0.010\;kg}} = 298302\frac{J}{kg}$\\\\
 
-$L_{1_{min}} =\displaystyle{\frac{\left(0.174\;kg \cdot 4182\frac{J}{kg \cdot K}\cdot 7.50^{\circ}C \right)-\left(0.184\;kg\cdot 4182 \frac{J}{kg \cdot K} \cdot 62.75^{\circ}C\right)}{0.184\;kg}} = 283331\frac{J}{kg}$\\\\
+$L_{f_{1_{min}}} =\displaystyle{\frac{\left(0.174\;kg \cdot 4182\frac{J}{kg \cdot K}\cdot 7.50^{\circ}C \right)-\left(0.010\;kg\cdot 4182 \frac{J}{kg \cdot K} \cdot 62.75^{\circ}C\right)}{0.010\;kg}} = 283331\frac{J}{kg}$\\\\
 
 $L_{f_1}=\displaystyle{\frac{L_{1_{max}}+L_{1_{min}}}{2}=\left(290816\pm 7486\right) \cdot \frac{J}{kg}=\left(2.9\pm 0.075\right)\cdot 10^{5} \frac{J}{kg}}$
 
 \subsection{Durchführung 2}
-$L_{f_2} = \displaystyle{\frac{m_{\rm{H_{2}O}} \cdot c_{\rm{H_{2}O}} \cdot \Delta\vartheta_2 - m_{\rm{Eis}} \cdot c_{\rm{H_2O}} \cdot \Delta\vartheta_1}{m_{\rm{Eis}}}}$
+
+$L_{f_2}=\displaystyle{\frac{\left(\left(m_3-m_{Kal}\right) \cdot 4182\frac{J}{kg \cdot K}\cdot \left(\vartheta_{T_3}-\vartheta_{T_4}\right)\right)-\left(\left(m_4-m_{3}\right)\cdot 4182 \frac{J}{kg \cdot K} \cdot \vartheta_{T_3}\right)}{m_4-m_{Kal}}} $\\\\
+
+$L_{f_{2_{max}}} =\displaystyle{\frac{\left(0.168\;kg \cdot 4182\frac{J}{kg \cdot K}\cdot 9.40^{\circ}C \right)-\left(0.015\;kg\cdot 4182 \frac{J}{kg \cdot K} \cdot 54.05^{\circ}C\right)}{0.184\;kg}} = 214244\frac{J}{kg}$\\\\
+
+$L_{f_{2_{min}}} =\displaystyle{\frac{\left(0.168\;kg \cdot 4182\frac{J}{kg \cdot K}\cdot 9.20^{\circ}C \right)-\left(0.015\;kg\cdot 4182 \frac{J}{kg \cdot K} \cdot 54.15^{\circ}C\right)}{0.184\;kg}} = 204458\frac{J}{kg}$\\\\
+
+$L_{f_2}=\displaystyle{\frac{L_{2_{max}}+L_{2_{min}}}{2}=\left(209351\pm 4893\right) \cdot \frac{J}{kg}=\left(2.1\pm 0.049\right)\cdot 10^{5} \frac{J}{kg}}$
+
 \section{Kommentar / Diskussion}
 
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