Answer:
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:
t = 1.4[s]
Explanation:
To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.
where:
P = impulse or lineal momentum [kg*m/s]
m = mass = 50 [kg]
v = velocity [m/s]
F = force = 200[N]
t = time = [s]
Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.
where:
m₁ = mass of the object = 50 [kg]
v₁ = velocity of the object before the impulse = 18.2 [m/s]
v₂ = velocity of the object after the impulse = 12.6 [m/s]