Question

Какое количество теплоты требуется для нагревания стальной детали массой 200 грамм от 35 до 1235°С. Правильный ответ 120кДж. Помогите пожалуйста на пишите дано, найти и решение.

Answer 1

Дано:
m=200 г = 0,2 кг
t₁=35 C
t₂=1235 С
c=500 Дж/кг*С - удельная теплоёмкость стали
Q=?
Q=cm(t₂-t₁)  Q=500*0,2*(1235-35)=120000 (Дж)= 120 (кДж)

Answer 2

Calculating Heat Required to Heat Steel

To calculate the heat required to heat a steel object, we can use the formula:

Q = mcΔT

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

Given: - Mass (m) = 200 grams = 0.2 kg - Initial temperature = 35°C - Final temperature = 1235°C - Specific heat capacity of steel (c) = 490 J/kg°C

Calculation

First, we need to convert the mass from grams to kilograms: - 200 grams = 0.2 kg

Next, we calculate the change in temperature: - ΔT = 1235°C - 35°C = 1200°C

Now, we can use the formula to calculate the heat energy (Q): - Q = (0.2 kg) * (490 J/kg°C) * (1200°C) - Q = 117,600 J = 117.6 kJ

So, the correct amount of heat required to heat the steel object from 35°C to 1235°C is 117.6 kJ.