Study the map extract provided on a scale of 1:50,000 and answer the questions that follow. In your new outline, insert and name: (i) the secondary road, (ii) the trigonometrical station, (iii) Mayo Jarandi settlement, (iv) River Juji and indicate its direction of flow with an arrow.

Step 1: Calculate the mass of cold water.
$m_c = 25~mL \times 1~\frac{g}{mL} = 25~g$

Step 2: Calculate the mass of hot water.
$m_h = 50~mL \times 1~\frac{g}{mL} = 50~g$

Step 3: Set up the heat balance equation (heat lost by hot water equals heat gained by cold water).
$m_h c (T_h - T_f) = m_c c (T_f - T_c)$
where $c = 1~\frac{cal}{g \cdot ^\circC}$, $T_h = 80^\circ\mathrm{C}$, $T_c = 20^\circ\mathrm{C}$.

Step 4: Substitute the known values ($c$ cancels out).
$50~g \left(80^\circC - T_f\right) = 25~g \left(T_f - 20^\circC\right)$

Step 5: Expand both sides.
$4000~cal - 50 T_f = 25 T_f - 500~cal$

Step 6: Add $50 T_f$ to both sides.
$4000~cal + 500~cal = 25 T_f + 50 T_f$
$4500~cal = 75 T_f$

Step 7: Solve for $T_f$.
$T_f = \frac{4500~cal}{75} = 60^\circC$

Step 8: Calculate heat lost by hot water.
$Q_h = m_h c (T_h - T_f) = 50~g \times 1~\frac{cal}{g \cdot ^\circC} \times (80^\circC - 60^\circC)$
$Q_h = 50~g \times 20^\circC = 1000~cal$

Step 9: Calculate heat gained by cold water.
$Q_c = m_c c (T_f - T_c) = 25~g \times 1~\frac{cal}{g \cdot ^\circC} \times (60^\circC - 20^\circC)$
$Q_c = 25~g \times 40^\circC = 1000~cal$

Final temperature of the mixture: $\boxed{60^\circC}$
Heat lost by hot water: $\boxed{1000~cal}$
Heat gained by cold water: $\boxed{1000~cal}$