Question

Яка кількість теплоти поглинеться при нагріванні 20кг сталі від 25 ⁰С до 80⁰С​

Answer 1

Q=c•m•t(tк-tн) вот в формулу подставь значения и все ( c найдёшь в таблице) а дельта t(80-25) вот и все

Answer 2

Ответ:

550кДж

Объяснение:

Дано:

с=500Дж/кг°С

m=20кг

∆t=80-25=55°C

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Q-?

Q=cm∆t=500*20*55=550000Дж

Answer 3

Heat Absorbed by Steel

To calculate the amount of heat absorbed by the steel when it is heated from 25°C to 80°C, we can use the formula:

Q = mcΔT

Where: - Q = heat absorbed (in joules) - m = mass of the steel (in kilograms) - c = specific heat capacity of steel (in joules per kilogram per degree Celsius) - ΔT = change in temperature (in degrees Celsius)

Calculation

Given: - m = 20 kg - Initial temperature (T1) = 25°C - Final temperature (T2) = 80°C

First, we need to find the specific heat capacity of steel. According to the search results, the specific heat capacity of steel is approximately 490 J/kg°C [[1]].

Now, we can calculate the heat absorbed using the formula:

Q = mcΔT

ΔT = T2 - T1 = 80°C - 25°C = 55°C

Q = 20 kg * 490 J/kg°C * 55°C

Q ≈ 539,000 joules

So, the amount of heat absorbed by 20 kg of steel when heated from 25°C to 80°C is approximately 539,000 joules.