Question

In the previous question you were provided with an extra piece of information: You were told that only liquid water would be present once equilibrium was established. If the final state of a system is not given, it is still possible to determine it from the initial temperatures and masses. The next few parts will help illustrate this point. Suppose that in an insulated container, 0.100 kg of water at 20.0∘C is mixed with 1.500kg of ice at −15.0∘C. You are asked to find the final temperature Tf of the system, but you are not told what the final phase of the system in equilibrium is.

Part H:
Question: Suppose the water is at its freezing point, 0∘C, but has not yet turned to ice. Find the amount of heat Qf released during the water's phase transition to ice. Your answer does not need to include a sign.

Answer 1

PART A:

The latent heat of fusion of water is 334 kJ/kg.

Specific heat capacity of ice is 2.108 kJ/kg.K.

Specific heat capacity of water is 4.187 kJ/kg.K.

The heat released by the 0.1 kg water to reach the temperature of 0 ^oC is

$Q_1=c_wm_w\Delta T=4.187\times 0.1\times (20-0)=8.374\ kJ$

The heat required by the 1.5 kg ice to reach the temperature of 0 ^oC is

$Q_2=c_im_i\Delta T=2.108\times 1.5\times (0-(-15))=47.43\ kJ$

The heat released by the 0.1 kg water to fuse to ice is

$Q_3=L_wm_w=334\times 0.1=33.4\ kJ$

Total heat released by the water to convert to the ice at 0 ^oC is

$Q=Q_1+Q_3=8.374+33.4=41.774\ kJ$

From the above results, we conclude that the phase of the equilibrium state would be solid (ice).

Now let us assume the final temperature at the equilibrium is T. That gives us

$Q+c_im_w(0-T)=c_im_i(T-(-15))$

$\Rightarrow 41.774+2.108\times 0.1\times (-T)=2.108\times 1.5\times (T+15)$

$\Rightarrow 41.774-0.2108T=3.162T+47.43$

$\Rightarrow 3.162T+0.2108T=41.774-47.43$

$\Rightarrow T=\frac{41.774-47.43}{3.162+0.2108}=-1.677\ ^oC$

So, the final temperature would be -1.677 ^oC.

PART H:

$Q_f=Lm=334\times 0.1=33.4\ kJ$

So, the answer is 33.4 kJ.