Respuesta :
Answer:
[tex]Q_{T}[/tex] = - 9.37 KJ
Explanation:
In this exercise the box has a caloric energy that must be eliminated. In several stages:
First you must go from the current temperature to zero centigrade (T = 0ºC) which is the freezing point of water, it is heat we will call it Q₁
Second you must lose heat without changing the temperature, it is changing state, latent heat [tex]Q_{L}[/tex]
Third, the ice temperature should be lowered from zero degrees to T = - (5 + B) = - (5 +6) = -11 ° C, we will call it Q₂
The heat removed is the sum of these heats
[tex]Q_{T}[/tex] = Q₁ + [tex]Q_{L}[/tex] + Q₂
Let's calculate each one. For this you have the two equations is calorimetry.
Q = m [tex]c_{e}[/tex] ([tex]T_{f}[/tex]-T₀)
Q = m L change of state
Let's calculate Q1, the data that give data is the mass m = 15.0 + A = 15.0 + 6 = 21.0 g, the initial temperature T₀ = 8.0 + C = 8.0 + 8 = 16ºC until the final temperature [tex]T_{f}[/tex] = 0ºC
Let's be careful because the specific heat is mixed in units, we must place the correct units so that they are simplified; this is the mass in grams, the temperature in degrees Kelvin
Q₁ = m [tex]c_{e}[/tex] ([tex]T_{f}[/tex]-T₀)
([tex]T_{f}[/tex]-T₀) = (273.15-289.15) = (0-16) = -16K
Q₁ = 21 4,186 (-16)
Q₁ = -1406.5 J
Let's calculate the latent heat
[tex]Q_{L}[/tex] = ±m L
[tex]Q_{L}[/tex] = -21 333
[tex]Q_{L}[/tex] = -6993 J
The negative sign was selected because heat is being lost
Calculate Q2
T = -11º C
Q₂ = 21 4,186 (-11 -0)
Q₂ = -966.97 J
The total heat is
[tex]Q_{T}[/tex] = -1406.5 - 6993 - 966.97
[tex]Q_{T}[/tex] = - 9366.5 J
The negative sign indicates that the heat is giving up.
Reduce to kJ
[tex]Q_{T}[/tex] = -9366.5 J (1kJ / 1000J)
[tex]Q_{T}[/tex] = - 9.37 KJ
