Question

Перед горячей штамповкой латунную болванку массой 3 кг нагрели от 15 градусов до 750 градусов. Какое кол-во теплоты получила болванка? Удельная теплоемкость латуни 380 Дж

Answer 1

Q=cm(t1-t2)=380Дж/кг°С•3кг•735°С=837900Дж

Answer 2

Calculation of Heat Absorbed by the Brass Blank

To calculate the amount of heat absorbed by the brass blank during the hot stamping process, we can use the formula:

Q = m * c * ΔT

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

Given: - Mass of the brass blank (m) = 3 kg - Initial temperature (T1) = 15 degrees Celsius - Final temperature (T2) = 750 degrees Celsius - Specific heat capacity of brass (c) = 380 J/kg°C

We can substitute these values into the formula to calculate the amount of heat absorbed:

Q = 3 kg * 380 J/kg°C * (750°C - 15°C)

Calculating the difference in temperature:

ΔT = T2 - T1 = 750°C - 15°C = 735°C

Substituting the values into the formula:

Q = 3 kg * 380 J/kg°C * 735°C

Calculating the result:

Q = 805,800 J

Therefore, the brass blank absorbed approximately 805,800 joules of heat during the hot stamping process.