Question

В воду массой 200 гр при температуре 20*С помещают стальную деталь массой 300 гр , имеющую температуру 10*С, и медную пластинку массой 400 гр при температуре 25*С.Найдите установившуюся температуру

Answer 1

Решение на фото, успехов :)

Answer 2

Calculation of the Final Temperature

To find the final temperature when a steel part with a mass of 300 g and a temperature of 10°C and a copper plate with a mass of 400 g and a temperature of 25°C are placed in water with a mass of 200 g and a temperature of 20°C, we can use the principle of energy conservation.

The principle of energy conservation states that the total energy of a closed system remains constant. In this case, the energy transferred between the water, steel part, and copper plate will result in the establishment of a final temperature.

Let's assume the final temperature is Tf.

The heat gained by the water can be calculated using the formula:

Q_water = m_water * c_water * (Tf - Tw)

Where: - Q_water is the heat gained by the water - m_water is the mass of the water (200 g) - c_water is the specific heat capacity of water (4.18 J/g°C) - Tf is the final temperature - Tw is the initial temperature of the water (20°C)

The heat lost by the steel part can be calculated using the formula:

Q_steel = m_steel * c_steel * (Tf - Ts)

Where: - Q_steel is the heat lost by the steel part - m_steel is the mass of the steel part (300 g) - c_steel is the specific heat capacity of steel (0.46 J/g°C) - Ts is the initial temperature of the steel part (10°C)

The heat lost by the copper plate can be calculated using the formula:

Q_copper = m_copper * c_copper * (Tf - Tc)

Where: - Q_copper is the heat lost by the copper plate - m_copper is the mass of the copper plate (400 g) - c_copper is the specific heat capacity of copper (0.39 J/g°C) - Tc is the initial temperature of the copper plate (25°C)

According to the principle of energy conservation, the heat gained by the water should be equal to the heat lost by the steel part and the copper plate:

Q_water = Q_steel + Q_copper

Substituting the formulas for Q_water, Q_steel, and Q_copper, we can solve for the final temperature Tf.

Let's calculate the final temperature using the given values and formulas.

Calculation:

Q_water = m_water * c_water * (Tf - Tw) Q_steel = m_steel * c_steel * (Tf - Ts) Q_copper = m_copper * c_copper * (Tf - Tc)

Q_water = Q_steel + Q_copper

Substituting the given values: - m_water = 200 g - c_water = 4.18 J/g°C - Tw = 20°C - m_steel = 300 g - c_steel = 0.46 J/g°C - Ts = 10°C - m_copper = 400 g - c_copper = 0.39 J/g°C - Tc = 25°C

We can now solve for Tf.

200 * 4.18 * (Tf - 20) = 300 * 0.46 * (Tf - 10) + 400 * 0.39 * (Tf - 25)

Simplifying the equation:

836(Tf - 20) = 138(Tf - 10) + 156(Tf - 25)

836Tf - 16720 = 138Tf - 1380 + 156Tf - 3900

Combining like terms:

836Tf - 138Tf - 156Tf = 16720 - 1380 + 3900

542Tf = 17140

Solving for Tf:

Tf = 17140 / 542 ≈ 31.66°C

Therefore, the final temperature when the steel part, copper plate, and water reach thermal equilibrium is approximately 31.66°C.

Please note that this calculation assumes that there is no heat loss to the surroundings and that the specific heat capacities remain constant over the given temperature range.