Question

Медную деталь, нагретую до 720°С, погрузили в 1,75 г воды при температуре 18°С. вся вода при этом нагрелась до 100°С и 75 г ее обратилось в пар. Определить массу детали. Потерями энергии пренебречь.

Answer 1

Дано:
tмеди=720
mводы1=1.75г
tводы1=18
tводы2=100
mводы2=75г
найти
mмед.дет.=?
решение:    Qводы=Q1+Q2
     Q1=4200*1.75*88=646.8 кДж
     Q2=2260*0.075=169.5 Дж
Qводы=646969.5Дж
Qводы=Qмед.дет
646969.5=400*mмед.дет*620
mмед.дет преблизительно= 2.6кг
ответ 2.6кг

Answer 2

Problem Analysis

We are given a copper part that is heated to 720°C and then submerged in 1.75 g of water at 18°C. The water heats up to 100°C, and 75 g of it turns into steam. We need to determine the mass of the copper part, assuming no energy losses.

Solution

To solve this problem, we can use the principle of conservation of energy. The heat gained by the water and the steam is equal to the heat lost by the copper part.

Let's calculate the heat gained by the water and the steam first.

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

Q_water = m_water * c_water * ΔT_water

Where: - Q_water is the heat gained by the water - m_water is the mass of the water - c_water is the specific heat capacity of water - ΔT_water is the change in temperature of the water

The specific heat capacity of water is approximately 4.18 J/g°C.

Substituting the given values: - m_water = 1.75 g - c_water = 4.18 J/g°C - ΔT_water = 100°C - 18°C = 82°C

Q_water = 1.75 g * 4.18 J/g°C * 82°C = 590.57 J

Next, let's calculate the heat gained by the steam. The heat gained by the steam can be calculated using the formula:

Q_steam = m_steam * c_steam * ΔT_steam

Where: - Q_steam is the heat gained by the steam - m_steam is the mass of the steam - c_steam is the specific heat capacity of steam - ΔT_steam is the change in temperature of the steam

The specific heat capacity of steam is approximately 2.03 J/g°C.

Substituting the given values: - m_steam = 75 g - c_steam = 2.03 J/g°C - ΔT_steam = 100°C - 100°C = 0°C (since the steam is already at its boiling point)

Q_steam = 75 g * 2.03 J/g°C * 0°C = 0 J

Now, let's calculate the heat lost by the copper part. The heat lost by the copper part can be calculated using the formula:

Q_copper = m_copper * c_copper * ΔT_copper

Where: - Q_copper is the heat lost by the copper part - m_copper is the mass of the copper part - c_copper is the specific heat capacity of copper - ΔT_copper is the change in temperature of the copper part

The specific heat capacity of copper is approximately 0.39 J/g°C.

Substituting the given values: - Q_copper = Q_water + Q_steam (since the heat lost by the copper part is equal to the heat gained by the water and the steam)

Q_copper = 590.57 J + 0 J = 590.57 J

Now, let's solve for the mass of the copper part.

Q_copper = m_copper * c_copper * ΔT_copper

Rearranging the equation:

m_copper = Q_copper / (c_copper * ΔT_copper)

Substituting the given values: - Q_copper = 590.57 J - c_copper = 0.39 J/g°C - ΔT_copper = 720°C - 100°C = 620°C

m_copper = 590.57 J / (0.39 J/g°C * 620°C) = 24.03 g

Therefore, the mass of the copper part is approximately 24.03 g.

Answer

The mass of the copper part is approximately 24.03 g.