Question

A=2cm; b=5cm; c=10cm; ρ=11.3•10^3 kg/m^3λ=2.5•10^4 Дж/кгc=140Дж/кг°Сt1=327°C t2=27°CНадо найти: Q. (Q=Q1+Q2), (Q=cm(t2-t1), Q=λm​

Answer 1

Ответ:

Q = 75,71 кДж

Объяснение:

a=0,02 m

b=0,05 m

c=0,1 m

ρ=11,3•10^3 kg/m^3

λ=2,5•10^4 Дж/кг

c=140 Дж/кг°С

t1=327°C

t2=27°C

Решение:

Количество теплоты, требуемое для нагревания свинца до температуры плавления: Q1 = c•m•(t2-t1)

Количество теплоты, требуемое для плавления свинца: Q2 = λ•m

Суммарное количество теплоты: Q = Q1 + Q2 = c•m•(t2-t1) + λ•m = (c•(t2-t1) + λ)•m

Масса свинца: m=ρ•V

Объем свинца: V=a•b•c

Значит, Q = (c•(t2-t1) + λ)•ρ•a•b•c = (140•(327-27) + 2,5•10^4)•11,3•10^3•0,02•0,05•0,1 = 75710(Дж) = 75,71(кДж)

Answer 2

To find the value of Q, we can use the given equations:

1. Q = Q1 + Q2
2. Q = cm(t2 - t1)
3. Q = λm

Given: A = 2 cm b = 5 cm c = 10 cm ρ = 11.3 x 10^3 kg/m^3 λ = 2.5 x 10^4 J/kg c = 140 J/(kg°C) t1 = 327°C t2 = 27°C

First, let's calculate the mass (m) using the density (ρ) and the volume (V) of the object.

V = A * b * c V = (2 cm) * (5 cm) * (10 cm) V = 100 cm^3 = 100 x 10^-6 m^3

m = ρ * V m = (11.3 x 10^3 kg/m^3) * (100 x 10^-6 m^3) m = 1.13 kg

Now, we can calculate Q using the equations mentioned above.

Using Equation 1: Q = Q1 + Q2

Q1 = cm(t2 - t1) Q1 = (1.13 kg) * (140 J/(kg°C)) * (27°C - 327°C) Q1 = (1.13 kg) * (140 J/(kg°C)) * (-300°C) Q1 = -38160 J

Q2 = λm Q2 = (2.5 x 10^4 J/kg) * (1.13 kg) Q2 = 28250 J

Q = Q1 + Q2 Q = -38160 J + 28250 J Q = -9910 J

Therefore, the value of Q is -9910 J. Note that the negative sign indicates that heat is being removed from the system.