Question

Какое количество теплоты отдаст стальная деталь массой 200 кг при остывании от 370 до 20 градусов удельная теплоемкость стали 500 джоулей

Answer 1

Дано :
m=200кг
t1=370°C
t2=20°C
c=500Дж
Найти:
Q-?
Решение:
Q=c*m*(t1-t2)
Q=500*200*(370-20)=100000*350=35000000Дж=35000кДж=35МДж
Ответ : 35МДж

Answer 2

Calculation of Heat Transfer

To calculate the amount of heat transferred by a steel component during cooling, we can use the formula:

Q = m * c * ΔT

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

In this case, we have: - m = 200 kg (mass of the steel component) - c = 500 J/kg°C (specific heat capacity of steel) - ΔT = 370°C - 20°C = 350°C (change in temperature)

Substituting these values into the formula, we can calculate the amount of heat transferred:

Q = 200 kg * 500 J/kg°C * 350°C

Let's calculate this:

Q = 200 kg * 500 J/kg°C * 350°C = 35,000,000 J

Therefore, the steel component will transfer 35,000,000 joules of heat during cooling from 370°C to 20°C.

Please note that the specific heat capacity value of 500 J/kg°C is taken from search result snippet