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Liono4ka [1.6K]

2 years ago

14

Dibuja la gráfica de calentamiento de un kilogramo de plomo que se encuentra inicialmente a 70ºC y pasa a una temperatura final

de 2000ºC, sabiendo que su temperatura de fusión es: 327.4ºC y la de ebullición es de 1725ºC

1 answer:

MariettaO [177]2 years ago

8 0

Answer:

Q= m $c_e$ ΔT and Q = m L

Explanation:

For this graph of temperature vs energy (heating) we must use two relations

* for when there is no change of state

Q= m $c_e$ ΔT

* for using there is change of state

Q = m L

the second expression is a consequence of the fact that all the energy supplied is used to change the state of the solid-liquid and liquid-gas system

the energy supplied is the sum of the energy in each interval

divide the system into intervals determined by the state change points

1) from T₀ = 70ºC to T_f = 327.4ºC, sample in solid-liquid state

c_e = 128 J / kg ºC

Q₁ = m c_e (T_f -To)

Q₁=1 128 (327.4 -70)

Q₁ = 3.29 10⁴ J

Q = Q₁ = 3.29 10⁴ J

2) when is it changing from solid to liquid

L = 2.45 10⁴ J / kg

Q2 = 1 2.45 10⁴

Q2 = 2.45 10⁴ J

Q = Q₁ + Q₂

Q = 5.74 10⁴ J

3) from to = 327.4ºC until T_f = 1725ºC, sample in liquid state

in the tables the specific heat of the solid and liquid state is the same

Q3 = m c_e (T_f -To)

Q3 = 1 128 (1725 -327.4)

Q3 = 1.79 10⁵ J

Q = Q₁ + Q₂ + Q₃

Q = (3.29 +2.45 + 17.9) 10⁴ J

Q = 23.64 10⁴ J

4) for when it is changing from the liquid state to the gaseous state

L_v = 8.70 10⁵ J / kg

Q₄ = m L_v

Q₄ = 1 8.70 10⁵

Q₄ = 8.70 10⁵ J

Q = Q₁ + Q₂ + Q₃ + Q₄

Q = (3.29 +5.74 + 17.9+ 87.0) 10⁴ J

Q = 110.64 10⁴ J

5) from To = 1725ºC to T_f = 2000ºC, sample in gaseous state

Q₅ = m c_e ΔT

Q₅ = 1 128 (2000 -1725)

Q₅ = 3.52 10⁴ J

Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅

Q = 114.16 104 J

the following table shows the points to be plotted

Energy (10⁴ J) Temperature (ºC)

0 70

3.29 327.4

5.74 327.4

23.64 1725

110.64 1725

114.16 2000

In the attachment we can see a graph of Temperature versus energy supplied

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