Question

5 литров кипятка взятые при температуре 100*С остывая отдали в окружающее пространство 1680 кДж энергии. Какой стала температура воды?

Answer 1

Могу предположить что вы можете воспользоваться такой формулой:
$Q=cm(t_0-t)=1680*10^3$

Answer 2

Calculation of the Final Temperature of Water

To determine the final temperature of the water after it has cooled down and released 1680 kJ of energy into the surrounding space, we can use the principle of energy conservation. The energy released by the water is equal to the energy absorbed by the surrounding space.

The specific heat capacity of water is approximately 4.18 J/g°C. We can use this value to calculate the change in temperature of the water.

Let's assume the initial temperature of the water is 100°C and the final temperature is T°C.

The energy released by the water can be calculated using the formula:

Energy released = mass of water × specific heat capacity of water × change in temperature

Given that the mass of water is 5 liters and the specific heat capacity of water is 4.18 J/g°C, we need to convert the mass of water from liters to grams.

1 liter of water has a mass of approximately 1000 grams.

Therefore, the mass of water is 5 × 1000 = 5000 grams.

Substituting the values into the formula, we have:

1680 kJ = 5000 g × 4.18 J/g°C × (100°C - T°C)

Now, let's solve for T°C:

1680 kJ = 5000 g × 4.18 J/g°C × (100°C - T°C)

Converting 1680 kJ to J:

1680 kJ = 1680 × 1000 J = 1,680,000 J

1,680,000 J = 5000 g × 4.18 J/g°C × (100°C - T°C)

Simplifying the equation:

1,680,000 J = 20,900 g°C × (100°C - T°C)

Dividing both sides of the equation by 20,900 g°C:

80.383 = 100 - T°C

Subtracting 80.383 from both sides of the equation:

T°C = 100 - 80.383

T°C ≈ 19.617

Therefore, the final temperature of the water is approximately 19.617°C.

Please note that this calculation assumes that there is no heat loss to the surroundings other than the 1680 kJ of energy released by the water.