Question

A (10.0+A) g ice cube at -15.0oC is placed in (125+B) g of water at 48.0oC.

Find the final temperature of the system when equilibrium is reached.

Ignore the heat capacity of the container and assume this is in a calorimeter, i.e. the system is thermally insulated from the surroundings.

Give your answer in oC with 3 significant figures.

Answer 1

Answer:

  $T_{f}$ = 38.0° C

Explanation:

The thermal transfer processes are governed by two equations one for there is a change of state, during this process the temperature does not change and another for which there is a change of temperature.  

     Q = ± m L  

     Q = m $c_{e}$ ΔT  

In calorimetric processes the heat transferred is equal to the heat absorbed by the system  

     -Qc = Qabs  

Let's apply these equations to our system, we will assume the quantity A and B zero, reduce the magnitudes to the SI system  

    m1 = 10 g = 0.010 kg  

     m2 = 125 g = 0.125 kg  

    $c_{e}$ (ice) = 2090 J / kg ºC  

    $c_{ew}$ (water) = 4186 J / kg ºC  

     Lf = 3.33 105 J / kg  

Let's write the expression for the heat given  

     Qc = m₂ $c_{ew}$ (T₀ - $T_{f}$ )  

     Qc = 0.125 4186 (48.0 - $T_{f}$)  

     Qc = 523.25 (48.0 - $T_{f}$)  

The expression for absorbed  is  

     Qabs = Q1 + Q2 + Q3  

calculate Q₁

     Q₁ = m $c_{e}$ ($T_{f}$ - T₀)  

     Q₁ = 0.010 2090 (0 - (-15.0))  

     Q₁ = 313.5 J  

Let's change state from ice to liquid water  

     Q₂ = + mL  

     Q₂ = 0.010 3.33 105  

     Q₂ = 3.33 103 J  

Now let's heat the liquid water to the equilibrium temperature

     Q₃ = m₁ $c_{ew}$ ($T_{f}$-T₀)  

     Q₃ = 0.01 4186 ($T_{f}$ - 0)  

     Q₃ = 41.86 $T_{f}$

let's calculate the maximum heat assigned

      Qc = 0.125 4186 (48.0 - 0)  

     Qc = 2.5 104 J  

     Qc> (Q₂ + Q₁)  

all the ice is melted

523.25 (48.0 - $T_{f}$) = Q1 + Q2 + 41.86 $T_{f}$

     (41.86 +523.25) $T_{f}$= 523.25 48 - Q1 -Q2  

     565.11 $T_{f}$ = 25116 - 313.5 - 3.33 103  

     $T_{f}$ = 21469.5 / 565.11  

$T_{f}$ = 38.0 ° C