Question

Лицар замовив у коваля меч із сталі. Для виконання замовлення коваль нагрів сталевий брус масою 13 кг до температури 500 °С, надав бруску потрібну форму та занурив у 17,5 кг води, щоб охолодити. Початкова температура води 20 °С. Визначте, яка температура води стала після встановлення теплової рівноваги.

Answer 1

Ответ:

56 °С

Объяснение:

m₁=13 кг

T₁=500 °С,

c₁=450  Дж/(кг*°С)

m₂=17,5 кг

T₂=20 °С

θ   ?   θ=T₁-ΔT₁=T₂+ΔT₂; ΔT₂=T₁-T₂-ΔT₁=500-20-ΔT₁=480-ΔT₁; Q₁=Q₂;

m₁*c₁*ΔT₁=m₂*c*ΔT₂=m₂*c*(480-ΔT₁); 13*450*ΔT₁=17,5*4200(480-ΔT₁);

ΔT₁=35280000/(73500+5850)=444 °С; θ=500-444=56 °С

Answer 2

Problem Analysis

In this problem, we are given that a knight ordered a sword made of steel from a blacksmith. The blacksmith heated a steel bar weighing 13 kg to a temperature of 500 °C and then shaped it into the required form. To cool down the steel bar, the blacksmith immersed it in 17.5 kg of water. We need to determine the final temperature of the water after thermal equilibrium is established.

Solution

To solve this problem, we can use the principle of conservation of energy. The energy gained by the water is equal to the energy lost by the steel bar.

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

Q_water = m_water * c_water * (T_final - T_initial)

where: - Q_water is the energy gained by the water - m_water is the mass of the water (17.5 kg) - c_water is the specific heat capacity of water (4.18 J/g°C) - T_final is the final temperature of the water (unknown) - T_initial is the initial temperature of the water (20 °C)

The energy lost by the steel bar can be calculated using the formula:

Q_steel = m_steel * c_steel * (T_initial - T_final)

where: - Q_steel is the energy lost by the steel bar - m_steel is the mass of the steel bar (13 kg) - c_steel is the specific heat capacity of steel (0.45 J/g°C) - T_initial is the initial temperature of the steel bar (500 °C) - T_final is the final temperature of the steel bar (unknown)

Since the energy gained by the water is equal to the energy lost by the steel bar, we can set up the following equation:

Q_water = Q_steel

Substituting the formulas for Q_water and Q_steel, we get:

m_water * c_water * (T_final - T_initial) = m_steel * c_steel * (T_initial - T_final)

Simplifying the equation, we can solve for T_final:

m_water * c_water * T_final - m_water * c_water * T_initial = m_steel * c_steel * T_initial - m_steel * c_steel * T_final

T_final * (m_water * c_water + m_steel * c_steel) = m_steel * c_steel * T_initial + m_water * c_water * T_initial

T_final = (m_steel * c_steel * T_initial + m_water * c_water * T_initial) / (m_water * c_water + m_steel * c_steel)

Now we can substitute the given values into the equation to find the final temperature of the water.

Calculation

Substituting the given values: - m_water = 17.5 kg - c_water = 4.18 J/g°C - T_initial = 20 °C - m_steel = 13 kg - c_steel = 0.45 J/g°C - T_initial = 500 °C

We can calculate the final temperature of the water using the formula:

T_final = (13 kg * 0.45 J/g°C * 500 °C + 17.5 kg * 4.18 J/g°C * 20 °C) / (17.5 kg * 4.18 J/g°C + 13 kg * 0.45 J/g°C)

Calculating the numerator: 13 kg * 0.45 J/g°C * 500 °C + 17.5 kg * 4.18 J/g°C * 20 °C = 2925 J + 1465 J = 4390 J

Calculating the denominator: 17.5 kg * 4.18 J/g°C + 13 kg * 0.45 J/g°C = 73.15 J/°C + 5.85 J/°C = 79 J/°C

Substituting the values into the formula: T_final = 4390 J / 79 J/°C ≈ 55.57 °C

Answer

Therefore, the final temperature of the water after thermal equilibrium is established is approximately 55.57 °C.