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

Q= m 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 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

answer: the rock sample is a sedimentary rock.

explanation:

it is given that:

mass of rock sample, m = 2.7 g

volume of rock sample,

so, density of rock sample will be

since, the density of rock sample is . hence, the rock sample is a sedimentary rock, whose densities varies between .