**Answer:**

= - 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

**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

= Q₁ + + Q₂

Let's calculate each one. For this you have the two equations is calorimetry.

Q = m (-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 = 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 (-T₀)

(-T₀) = (273.15-289.15) = (0-16) = -16K

Q₁ = 21 4,186 (-16)

Q₁ = -1406.5 J

Let's calculate the latent heat

= ±m L

= -21 333

= -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

= -1406.5 - 6993 - 966.97

= - 9366.5 J

The negative sign indicates that the heat is giving up.

Reduce to kJ

= -9366.5 J (1kJ / 1000J)

= - 9.37 KJ