Question

A container with (15.0 + A) g of water at (8.0 + C) oC is placed in a freezer. How much heat must be removed from the water to turn it to ice at –(5.0 + B) oC? Ignore the heat capacity of the container. Give your answer in kilo-joules (kJ) with 3 significant figures.
Specific heat of ice: 2.090 J/g K
Specific heat of water: 4.186 J/g K
Latent heat of fusion for water: 333 J/g
A=6 B=6 C=8

Answer 1

Answer:

$Q_{T}$ = - 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 $Q_{L}$

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  

    $Q_{T}$ = Q₁ + $Q_{L}$ + Q₂  

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

     Q = m $c_{e}$ ($T_{f}$-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 $T_{f}$ = 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 $c_{e}$ ($T_{f}$-T₀)  

    ($T_{f}$-T₀) = (273.15-289.15) = (0-16) = -16K  

    Q₁ = 21 4,186 (-16)  

     Q₁ = -1406.5 J  

Let's calculate the latent heat  

    $Q_{L}$ = ±m L  

    $Q_{L}$ = -21 333  

   $Q_{L}$ = -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  

   $Q_{T}$ = -1406.5 - 6993 - 966.97  

   $Q_{T}$ = - 9366.5 J  

The negative sign indicates that the heat is giving up.  

Reduce to kJ  

$Q_{T}$ = -9366.5 J (1kJ / 1000J)  

$Q_{T}$ = - 9.37 KJ