Question

Задача.1.До якої температури був нагрітий чавунний брусок масою 1.8 кг, якщо при його охолодженні до 20 °С виділилося 77,76 кДж теплової енергії? Питома теплоємність чавуну 550 Дж/(кг-°С). Задача.2.Визначте кiлькiсть теплоти, необхідну для нагрівання від 15 до 20 °C повітря в кімнаті, Довжина якої 5 м, ширина 4 м та висота 2,8 м.Задача.3.У чавунну ванну, маса якої 200 кг, спочатку налили воду температурою 20 °С, а потім додали води температурою 80 ºС. Після встановлення Тепловоï рівноваги температура води стала дорівнювати 60 °С. Яку масу холодної та гарячої води налили у ванну, якщо загальний об'єм води у ванні склав 160 л?

Answer 1

Ответ:

Объяснение:

Задача 1

77,76 кДж = 77 760 Дж

T-20° = 77760/ 1.8 * 550 = 77760/990 = 78.5° - на столько охладили.

T = 78.5+20 = 98.5° - нач температура

2) теплоемкость воздуха около 1000 Дж

Объем комнаты 4*5*2,8 = 20*2,8 = 56 м3 - объем воздуха.

плотность вохдуха 1,2 кг/м3

масса = 1,2*56 = 67,2 кг.

Q = 1000 Дж *67,2* 5 = 1000 *336 = 336 000 Дж = 336 кДж

3) 60° = 20° Х +80° Y / 160

X - холодная вода

Y - горячая вода

X + Y = 160 литров

20 Х +80 Y = 160*60

20 X +80 Y = 9600

X = 160- Y

20 (160-Y) +80 Y = 9600

3200 -20 Y +80 Y = 9600

60 Y = 9600-3200

60 Y = 6400

Y = 106.6 л - горячая вода

X = 160-106.6 = 53.4 л - холодная вода

Answer 2

Problem 1: Heating a Cast Iron Block

To find the temperature to which a cast iron block weighing 1.8 kg was heated, given that 77.76 kJ of thermal energy was released during its cooling to 20 °C, we can use the specific heat capacity of cast iron.

The specific heat capacity of cast iron is given as 550 J/(kg-°C) [[1]].

Let's assume the initial temperature of the cast iron block is T °C.

The thermal energy released during cooling can be calculated using the formula:

Q = m * c * ΔT

Where: - Q is the thermal energy released (77.76 kJ or 77,760 J), - m is the mass of the cast iron block (1.8 kg), - c is the specific heat capacity of cast iron (550 J/(kg-°C)), - ΔT is the change in temperature (T - 20 °C).

Substituting the given values into the formula, we can solve for T:

77,760 = 1.8 * 550 * (T - 20)

Simplifying the equation:

77,760 = 990 * (T - 20)

77,760 = 990T - 19,800

990T = 97,560

T ≈ 98.67 °C

Therefore, the cast iron block was heated to approximately 98.67 °C before being cooled down to 20 °C.

[[1]] Source: [Engineering Toolbox - Specific Heat Capacity of Metals Table](https://www.engineeringtoolbox.com/specific-heat-capacity-d_391.html)

Problem 2: Heating the Air in a Room

To determine the amount of heat required to heat the air in a room from 15 °C to 20 °C, we can use the formula:

Q = m * c * ΔT

Where: - Q is the amount of heat required, - m is the mass of the air, - c is the specific heat capacity of air, - ΔT is the change in temperature.

To find the mass of the air, we need to calculate the volume of the room:

Volume = length * width * height

Given: - Length = 5 m - Width = 4 m - Height = 2.8 m

Volume = 5 * 4 * 2.8 = 56 m³

The density of air at room temperature and pressure is approximately 1.225 kg/m³ [[2]].

Therefore, the mass of the air in the room is:

Mass = Volume * Density = 56 * 1.225 = 68.6 kg

The specific heat capacity of air is approximately 1005 J/(kg-°C) [[3]].

Substituting the values into the formula:

Q = 68.6 * 1005 * (20 - 15) = 343,215 J

Therefore, approximately 343,215 J of heat is required to heat the air in the room from 15 °C to 20 °C.

[[2]] Source: [Engineering Toolbox - Air - Density and Specific Weight](https://www.engineeringtoolbox.com/air-density-specific-weight-d_600.html) [[3]] Source: [Engineering Toolbox - Specific Heat Capacity of Gases](https://www.engineeringtoolbox.com/specific-heat-capacity-gases-d_1592.html)

Problem 3: Mixing Hot and Cold Water in a Cast Iron Bathtub

To determine the mass of hot and cold water poured into a cast iron bathtub, given that the total volume of water in the bathtub is 160 liters and the temperature of the water reaches equilibrium at 60 °C, we can use the principle of conservation of energy.

Let's assume the mass of the cold water poured into the bathtub is m_cold kg and the mass of the hot water poured into the bathtub is m_hot kg.

The specific heat capacity of water is approximately 4186 J/(kg-°C) [[4]].

Using the principle of conservation of energy, the heat gained by the cold water is equal to the heat lost by the hot water:

m_cold * c * (60 - 20) = m_hot * c * (60 - 80)

Simplifying the equation:

m_cold * 40 = m_hot * (-20)

m_cold = -0.5 * m_hot

The total volume of water in the bathtub is given as 160 liters, which is equal to 160 kg since the density of water is approximately 1000 kg/m³.

Therefore, we can write the equation:

m_cold + m_hot = 160

Substituting the value of m_cold from the previous equation:

-0.5 * m_hot + m_hot = 160

0.5 * m_hot = 160

m_hot ≈ 320 kg

Substituting the value of m_hot into the equation:

m_cold = -0.5 * 320 ≈ -160 kg

Since mass cannot be negative, we can ignore the negative sign and consider the mass of the cold water as approximately 160 kg.

Therefore, approximately 160 kg of cold water and 320 kg of hot water were poured into the cast iron bathtub.

[[4]] Source: [Engineering Toolbox - Specific Heat Capacity of Liquids](https://www.engineeringtoolbox.com/specific-heat-capacity-liquids-d_151.html)