Question

решите задачи по физике,подробно,записав все,что дано и с полным решением1.В воду с темпиратурой 20*с влили ртуть,масса которой равна массе воды.Определите начальную темпиратуру ртути,если установившаяся темпиратура стала 21*с2.сколько граммов древестного угля надо сжечь в самоваре,емкость которого 5 л,чтобы нагреть в нём воду от 20 до 100*с?Учесть,что только 25% выделяемой энергии расходуется на нагревание.3.Чтобы охладить до 60*с 2 л воды,взятой при темпиратуре 80*с,в неё добавляют холодную воду,темпиратура которой 10*с.Сколько литров холодной воды требуется добавить?

Answer 1

3) Дано:                     Q1=Q2     С-удельная теплоемкость воды
t1=80 градусов          С-удельная теплоемкость воды
t2=10 градусов          Q1=C*m1*(t1-t)=C*po(плотность воды)*V1*(t1-t)
t=60 градусов            Q2=C*m2*(t-t2)=C*po(плотность воды)*V2*(t-t2)
t2=10 градусов        C*po(плотность воды)*V1*(t1-t)=C*po(плотность воды)*V2*(t-t2)      V1=2 л                      V1*(t1-t)=V2*(t-t2)
  V2=?                         V2=V1*(t1-t)\(t-t2)=2*(80-60)\(60-10)=0,8[л]

Answer 2

Task 1: Determining the Initial Temperature of Mercury

Given: - Initial temperature of water, T1 = 20°C - Mass of mercury = mass of water - Final temperature after equilibrium, T2 = 21°C

To find: - Initial temperature of mercury

Let's calculate the initial temperature of mercury using the given information.

The specific heat capacity of water, c_water = 4.18 J/g°C The specific heat capacity of mercury, c_mercury = 0.14 J/g°C

The heat gained by the mercury equals the heat lost by the water: m_water * c_water * (T2 - T1) = m_mercury * c_mercury * (T_mercury - T1)

Since the mass of mercury is equal to the mass of water, we can simplify the equation to: c_water * (T2 - T1) = c_mercury * (T_mercury - T1)

Substitute the given values: 4.18 * (21 - 20) = 0.14 * (T_mercury - 20)

Solving for T_mercury: 4.18 = 0.14 * (T_mercury - 20) 4.18 = 0.14T_mercury - 2.8 0.14T_mercury = 4.18 + 2.8 0.14T_mercury = 6.98 T_mercury = 6.98 / 0.14 T_mercury = 49.86°C

The initial temperature of the mercury is 49.86°C [[1]]

Task 2: Amount of Charcoal Needed to Heat Water

Given: - Volume of the samovar, V = 5 L - Initial temperature of water, T1 = 20°C - Final temperature of water, T2 = 100°C - Efficiency of energy utilization, η = 25%

To find: - Mass of charcoal needed to heat the water

The energy required to heat the water can be calculated using the formula: Q = mcΔT

Where: Q = Heat energy required m = Mass of water c = Specific heat capacity of water ΔT = Change in temperature

The energy required to heat the water from T1 to T2 is given by: Q = mcΔT = 1000g * 4.18J/g°C * (100 - 20)°C = 334,400J

Considering the efficiency, the actual energy required is: Q_actual = Q / η = 334,400J / 0.25 = 1,337,600J

The energy produced by burning the charcoal can be calculated using its calorific value: 1g of charcoal produces approximately 29,000J of energy

The mass of charcoal needed is: m_charcoal = Q_actual / 29,000J/g = 1,337,600J / 29,000J/g ≈ 46g

Approximately 46 grams of charcoal are needed to heat the water [[2]]

Task 3: Adding Cold Water to Cool the Water

Given: - Initial volume of hot water, V1 = 2 L - Initial temperature of hot water, T1 = 80°C - Final temperature after cooling, T2 = 60°C - Temperature of cold water, T_cold = 10°C

To find: - Volume of cold water needed to be added

The heat lost by the hot water equals the heat gained by the cold water: m_hot * c_water * (T1 - T2) = m_cold * c_water * (T2 - T_cold)

Solving for m_cold: m_cold = (m_hot * (T1 - T2)) / (T2 - T_cold) m_cold = (2 * 4.18 * (80 - 60)) / (60 - 10) m_cold = (2 * 4.18 * 20) / 50 m_cold = 167.2 / 50 m_cold = 3.344 L

Approximately 3.344 liters of cold water are required to be added [[3]]