Sure, let's break down this problem step-by-step.
1. Identify the given information:
- Volume of the first quantity of water, [tex]\( V_1 = 20 \)[/tex] litres.
- Temperature of the first quantity of water, [tex]\( T_1 = 80^\circ C \)[/tex].
- Volume of the second quantity of water, [tex]\( V_2 = 30 \)[/tex] litres.
- Temperature of the second quantity of water, [tex]\( T_2 = 20^\circ C \)[/tex].
2. Calculate the total volume of the mixed water:
The total volume [tex]\( V_\text{total} \)[/tex] is the sum of the volumes of the two quantities.
[tex]\[
V_\text{total} = V_1 + V_2 = 20 \text{ litres} + 30 \text{ litres} = 50 \text{ litres}
\][/tex]
3. Determine the method to find the final temperature of the mixture:
To find the final temperature [tex]\( T_\text{final} \)[/tex] of the mixture, we use the principle of conservation of energy, which tells us that the heat gained by the cooler water will be equal to the heat lost by the warmer water when they reach thermal equilibrium. This can be done using a weighted average based on the volumes and temperatures of the two water quantities.
4. Set up the formula for the weighted average:
[tex]\[
T_\text{final} = \frac{(V_1 \cdot T_1) + (V_2 \cdot T_2)}{V_\text{total}}
\][/tex]
5. Plug in the given values into the formula:
[tex]\[
T_\text{final} = \frac{(20 \text{ litres} \cdot 80^\circ C) + (30 \text{ litres} \cdot 20^\circ C)}{50 \text{ litres}}
\][/tex]
6. Calculate the numerator:
[tex]\[
(20 \cdot 80) + (30 \cdot 20) = 1600 + 600 = 2200
\][/tex]
7. Divide by the total volume to find the final temperature:
[tex]\[
T_\text{final} = \frac{2200}{50} = 44.0^\circ C
\][/tex]
8. Conclusion:
The final temperature of the mixture is [tex]\( 44.0^\circ C \)[/tex].
So, when 20 litres of water at [tex]\( 80^\circ C \)[/tex] is mixed with 30 litres of water at [tex]\( 20^\circ C \)[/tex], the final temperature of the mixture is [tex]\( 44.0^\circ C \)[/tex].
