To 5 liters of water with a temperature of 90 degrees, 10 liters of water were added, after which the water

To 5 liters of water with a temperature of 90 degrees, 10 liters of water were added, after which the water temperature dropped to 40 degrees. What was the temperature of the added water?

1. Let:

V1 = 5 l;
t1 = 90 °;
V2 = 10 l;
t2 =?;
t = 40 °.
2. Let’s compose the equations for volumes and energies:

{V1 + V2 = V;
{t1V1 + t2V2 = tV;

{V1 + V2 = V;
{t2V2 = tV – t1V1;
{V1 + V2 = V;
{t2 = (tV – t1V1) / V2;
{V1 + V2 = V;
{t2 = (t (V1 + V2) – t1V1) / V2.
3. Substitute the values of the variables:

t2 = (t (V1 + V2) – t1V1) / V2 = (40 (5 + 10) – 90 * 5) / 10 = (40 * 15 – 90 * 5) / 10 = (600 – 450) / 10 = 150/10 = 15 (°).

Answer: 15 °.