Answer: $68^{\circ} \mathrm{C}$.

Solution. The heat balance equation in the first case is $c_{\mathrm{B}} m_{\mathrm{B}} 20=c_{\text {b }} m_{\text {b }} 60$. The heat balance equation in the second case:

$$
c_{\mathrm{B}} m_{\mathrm{B}}(80-t)+c_{6} m_{6}(80-t)=c_{6} m_{6}(t-20) .
$$

We get $c_{6} m_{6} \cdot 3 \cdot(80-t)+c_{6} m_{6}(80-t)=c_{6} m_{6}(t-20)$. As a result: $t=68^{\circ} \mathrm{C}$.
6. (15 points) A metal bar with a temperature of $20^{\circ} \mathrm{C}$ was submerged into water taken at a temperature of $100^{\circ} \mathrm{C}$. After thermal equilibrium was established, the temperature was found to be $80^{\circ} \mathrm{C}$. After this, without removing the first bar from the water, another identical bar with a temperature of $20^{\circ} \mathrm{C}$ was submerged into the water. What will the temperature of the water be after thermal equilibrium is established?