To find the temperature $t$ of the hot water, we can use the principle of conservation of energy. The heat lost by the hot water must equal the heat gained by the cooler water.

Let:

• $m$ be the mass of the hot water (with temperature $t$).
• $2m$ be the mass of the cooler water (with temperature $30^\circ C$).
• $T_f$ be the final temperature of the mixture, which is $50^\circ C$.

Using the formula for heat transfer, we have:

$\text{Heat lost by hot water} = \text{Heat gained by cooler water}$

This can be expressed as:

$m c (t - T_f) = 2m c (T_f - 30)$

Where $c$ is the specific heat capacity of water. Since $c$ and $m$ are common to both sides, they can be canceled out:

$t - 50 = 2(50 - 30)$

Calculating the right-hand side:

$t - 50 = 2 \times 20 = 40$

Now, solving for $t$:

$t = 50 + 40 = 90$

Thus, the temperature $t$ of the hot water is 90 degrees Celsius.

If you want further details or have any questions, let me know! Here are some related questions you might consider:

1. What assumptions are made in this calculation?
2. How would the result change if the final temperature were different?
3. What is the effect of using a different amount of cooler water?
4. How does the specific heat capacity of water influence the result?
5. Can this principle be applied to other substances besides water?

Tip: Always ensure to account for the masses and temperatures of all substances involved when using conservation of energy principles!