Question

решить задачи: 1.Какое количество теплоты получит 3 л воды при нагревании от 20 до 85 град., 2.Какое количество теплоты отдаст чугунная плита размерами 3х1,5х2 м, если она остыла от 120 до 50 град?

Answer 1

1) Дано

V=3л; t1=20 °C;t2=85 °C;c=4200Дж/кг*°C ;ρ=1000кг/$m^{3}$;

Реш:

Q=cm(t2-t1)

m=Vρ=3*$10^{-3}$*$10^{3}$=3кг

Q=4200*3*(85-20)=819000Дж=819кДж

2) Дано

a=3м;b=1.5м;d=2м; t1=120°С; t2=50°С; с=540Дж/кг*°C ;ρ=7000кг/$m^{3}$;

реш:

Q=cm(t1-t2)

V=9м³; m=Vρ =9*7000=63000кг;

Q=540*63000*70=2,3814*$10^{9}$Дж

Answer 2

Heat Calculation for Water and Iron Plate

1. Heat Calculation for Water:

To calculate the amount of heat gained by water when heated from 20°C to 85°C, we can use the formula:

\[ Q = mc\Delta T \]

Where: - \( Q \) = heat energy - \( m \) = mass of the substance - \( c \) = specific heat capacity of the substance - \( \Delta T \) = change in temperature

Given: - \( m = 3 \, \text{kg} \) (since the density of water is \( 1 \, \text{kg/L} \)) - \( c = 4.18 \, \text{J/g°C} \) (specific heat capacity of water) - \( \Delta T = 85°C - 20°C = 65°C \)

Substitute the values into the formula: \[ Q = 3 \, \text{kg} \times 4.18 \, \text{J/g°C} \times 65°C \]

Calculating the value: \[ Q = 3 \, \text{kg} \times 4.18 \, \text{J/g°C} \times 65°C = 785.7 \, \text{kJ} \]

So, the amount of heat gained by 3 liters of water when heated from 20°C to 85°C is 785.7 kJ.

2. Heat Calculation for Iron Plate:

To calculate the amount of heat lost by the iron plate when cooled from 120°C to 50°C, we can use the same formula:

\[ Q = mc\Delta T \]

Where: - \( Q \) = heat energy - \( m \) = mass of the substance - \( c \) = specific heat capacity of the substance - \( \Delta T \) = change in temperature

Given: - Dimensions: \( 3 \, \text{m} \times 1.5 \, \text{m} \times 2 \, \text{m} \) (assuming the plate is a rectangular prism) - Density of iron: \( 7.87 \, \text{g/cm}^3 \) - Specific heat capacity of iron: \( 0.45 \, \text{J/g°C} \)

First, calculate the mass of the iron plate: \[ \text{Volume} = 3 \, \text{m} \times 1.5 \, \text{m} \times 2 \, \text{m} = 9 \, \text{m}^3 = 9000 \, \text{L} \] \[ \text{Mass} = 9000 \, \text{L} \times 7.87 \, \text{g/L} = 70,830 \, \text{kg} \]

Now, calculate the heat lost: \[ \Delta T = 120°C - 50°C = 70°C \] \[ Q = 70,830 \, \text{kg} \times 0.45 \, \text{J/g°C} \times 70°C = 2,480,550 \, \text{kJ} \]

So, the amount of heat lost by the iron plate when cooled from 120°C to 50°C is 2,480,550 kJ.