Question

СРОЧНООООООООООООООО ДАЮ 50 баллов У ванну налили 100л води при температурі 120С, після чого долили 50л гарячої води при температурі 860С. Чому дорівнює температура суміші? Теплоту, яка була витрачена на нагрівання ванни, не враховувати.(питома теплоємність води 4200 Дж/(кг ·0C), густина води 1000 кг/м3)

Answer 1

Ответ:

≈36.7°

Объяснение:

V1=100*10⁻³ м³   V2=50*10⁻³ м³    t1=12°   t2=86°    tc=?

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m=p*V

Q1=Q2

c*m1*(tc-t1)=c*m2*(t2-tc)

V1*tc-V1*t1=V2*t2-V2*tc

tc*(V1+V2)=V2*t2+V1*t1

tc=(V2*t2+V1*t1)/(V1+V2)=(50*86+100*12)/(100+50)≈36.7°

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Answer 2

Ответ:

366,7°Сельсия

Объяснение:

Смотри на фото.

Answer 3

Calculation of the Final Temperature of the Mixture

To calculate the final temperature of the mixture, we can use the principle of conservation of energy. The total heat gained by the cold water is equal to the total heat lost by the hot water. We can express this as:

Heat gained by cold water = Heat lost by hot water

The heat gained or lost by a substance can be calculated using the formula:

Q = m * c * ΔT

Where: - Q is the heat gained or lost (in Joules) - m is the mass of the substance (in kilograms) - c is the specific heat capacity of the substance (in Joules per kilogram per degree Celsius) - ΔT is the change in temperature (in degrees Celsius)

Let's calculate the final temperature of the mixture step by step.

Step 1: Calculating the heat lost by the hot water

The initial temperature of the hot water is 86°C, and the final temperature is unknown. The mass of the hot water can be calculated using the density of water and the volume of water added.

Given: - Initial temperature of hot water (T1) = 86°C - Mass of hot water (m1) = 50 kg (since 1 liter of water has a mass of 1 kg) - Specific heat capacity of water (c) = 4200 J/(kg·°C)

Using the formula Q = m * c * ΔT, we can calculate the heat lost by the hot water:

Q1 = m1 * c * (T1 - Tf)

Step 2: Calculating the heat gained by the cold water

The initial temperature of the cold water is 12°C, and the final temperature is unknown. The mass of the cold water is given as 100 kg.

Given: - Initial temperature of cold water (T2) = 12°C - Mass of cold water (m2) = 100 kg - Specific heat capacity of water (c) = 4200 J/(kg·°C)

Using the formula Q = m * c * ΔT, we can calculate the heat gained by the cold water:

Q2 = m2 * c * (Tf - T2)

Step 3: Equating the heat gained and lost

Since the total heat gained by the cold water is equal to the total heat lost by the hot water, we can equate the two equations:

Q1 = Q2

m1 * c * (T1 - Tf) = m2 * c * (Tf - T2)

Step 4: Solving for the final temperature (Tf)

Now we can solve the equation for the final temperature (Tf):

m1 * c * (T1 - Tf) = m2 * c * (Tf - T2)

Substituting the given values: - m1 = 50 kg - T1 = 86°C - m2 = 100 kg - T2 = 12°C - c = 4200 J/(kg·°C)

We can simplify the equation and solve for Tf.

Please wait a moment while I calculate the final temperature.