HTX Frederikshavn & Hjørring

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===== Energibevarelse =====

==== Ø2.16 ====
  * $m_{k} = 1,25 kg$
  * $m_{v} = 0,75 kg$
  * $t_{k\_start} = 18,0 ^\circ C$
  * $t_{v\_start} = 72,0 ^\circ C$
  * $c_{vand} = 4180 \frac{J}{kg \times K}$

\begin{align}
Q_1 &+ Q_2 = 0\\
m_1 \times c_1 \times \Delta t_1 &+ m_2 \times c_2 \times \Delta t_2 = 0 \\
m_1 \times c_1 \times (t_{f} - t_{1\_start}) &+ m_2 \times c_2 \times (t_{f} - t_{2\_start}) = 0 \\
\end{align}


$$ 1,25 kg \times 4180 \times (t_f - 18) + 0,75 kg \times 4180 \times (t_f - 72) = 0 $$

\begin{align}
1,25 kg \times 4180 \times (t_f - 18) + 0,75 kg \times 4180 \times (t_f - 72) &= 0 \\
5225(t_f - 18) + 3135(t_f - 72) &= 0 \\
5225t_f - 94050 + 3135t_f - 225720 &= 0 \\
5225t_f + 3135t_f &= 94050 + 225720 \\
8360t_f &= 319770 \\
t_f &= \frac{319770}{8360} \\
t_f &= 38,25 ^\circ C
\end{align}

----

==== Ø2.17 ====

  * $m_{alu} = 1,4~kg$
  * $c_{alu} = 900 \frac{J}{kg}$
  * $t_{alu\_start} = 19 ^\circ C$

  * $m_{zink} = 2,3~kg$
  * $c_{zink} = 389 \frac{J}{kg}$
  * $t_{zink\_start} = 75 ^\circ C$

Vi antager at den varmeisolerede læderpose er et isoleret system. dvs $ \Delta E_{indre} = 0 $

\begin{align}
Q_{alu} &+ Q_{zink} = 0\\
m_{alu} \times c_{alu} \times \Delta t_{alu} &+ m_{zink} \times c_{zink} \times \Delta t_{zink} = 0 \\
m_{alu} \times c_{alu} \times (\mathbf{t_{fælles}} - t_{alu\_start}) &+ m_{zink} \times c_{zink} \times (\mathbf{t_{fælles}} - t_{zink\_start}) = 0 \\
\end{align}

\begin{align}
1,4 kg \times 900 \times (t_f - 19) + 2,3 kg \times 389 \times (t_f - 75) &= 0 \\
1260(t_f - 19) + 894,7(t_f - 75) &= 0 \\
1260t_f - 23940 + 894,7t_f - 67102,5 &= 0 \\
2154,7t_f &= 91042,5\\
t_f &= \frac{91042,5}{2154,7}\\
t_f &\approx 42,25 ^\circ C
\end{align}

----
==== Ø2.18 ====


  * $t_{fælles} = 19,7 ^\circ C$

  * $m_{vand} = 0,5~kg$
  * $c_{vand} = 4180 \frac{J}{kg}$
  * $t_{vand\_start} = 16 ^\circ C$
  * $\Delta t_{vand} = (19,7 - 16) = 3,5 ^\circ C$

  * $m_{lod} = 0,1~kg$
  * $c_{lod} = ??~ \frac{J}{kg}$
  * $t_{lod\_start} = 100^\circ C$
  * $\Delta t_{lod} = (19,7 - 100) = -80,5 ^\circ C$


\begin{align}
Q_{vand} &+ Q_{lod} = 0\\
m_{vand} \times c_{vand} \times \Delta t_{vand} &+ m_{lod} \times c_{lod} \times \Delta t_{lod} = 0 \\
\end{align}

\begin{align}
0,5 kg \times 4180 \times 3,5 ^\circ C + 0,1 kg \times c_{lod} \times -80,5 ^\circ C &= 0 \\
0,1 kg \times c_{lod} \times -80,5 ^\circ C &= - 0,5 kg \times 4180 \times 3,5 ^\circ C\\
c_{lod} &= \frac{- 0,5 kg \times 4180 \times 3,5 ^\circ C}{0,1 kg \times -80,5 ^\circ C} \\
c_{lod} &= \frac{- 0,5 kg \times 4180 \times 3,5 ^\circ C}{0,1 kg \times-80,5 ^\circ C} \\
c_{lod} &= \frac{-7315}{-8,05}\\
c_{lod} &= 908,7~\frac{J}{kg}
\end{align}

Ved opslag på specifikvarmekapacitet Orbit-B side 44. ses at loddet formentlig er lavet af **aluminium**.

----

==== Ø2.19 ====

  * $m_{vand} = 0,600~kg$
  * $c_{vand} = 4180 \frac{J}{kg \times ^\circ C}$
  * $t_{vand\_start} = 15 ^\circ C$

  * $m_{bly} = 0,100~kg$
  * $c_{bly} = 130~\frac{J}{kg \times ^\circ C}$
  * $t_{bly\_start} = 100^\circ C$

\begin{align}
Q_{vand} &+ Q_{bly} = 0\\
m_{vand} \times c_{vand} \times \Delta t_{vand} &+ m_{bly} \times c_{bly} \times \Delta t_{bly} = 0 \\
m_{vand} \times c_{vand} \times (t_f - t_{vand\_start}) &+ m_{bly} \times c_{bly} \times (t_f - t_{bly\_start}) = 0 \\
\end{align}

\begin{align}
0,600 \times 4180 \times (t_f - 15) + 0,400 \times 130 \times (t_f - 100) &= 0\\
2508t_f - 37620 + 52t_f - 5200 &= 0\\
2560t_f &= 42820\\
t_f &= \frac{42820}{2560} \\
t_f &= 16,73 ^\circ C 
\end{align}