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

\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}

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

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

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}

• $t_{fælles} = 19,7 ^\circ C$

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

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

\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.

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

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

\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}