Ice cube in water: entropy calculation (Calorimetry)

[Su6had1p]

Homework Statement

A 50 g cube of ice at -10 deg Celsius is dropped in a container filled with 10 g water at 10 deg Celsius. Calculate the change in entropy of the system assuming no heat exchange with the surroundings.

Relevant Equations

Specific heat of ice = 0.5 cal/g/K
Specific heat of water = 1 cal/g/K
Latent heat of fusion = 80 cal/g
1 cal = 4.18 J

Since I don't know what the system will look like in the end I tried to check for the extreme cases.
Case 1 if the whole ice melts :
$$T_f = 170.5 K$$
Case 2 if the whole water freezes :
$$T_f = 291.33 K$$
Both of which are unreasonable. Therefore the mixture contains both water and ice, this can only happen if the temperature is zero degree celsius.
Suppose $x$ amount of ice melts and $(50 - x)$ amount of ice remains
the heat equation can be written as :
$$m_w C_w \Delta T = m_{ice} C_{ice} \Delta T$$
$$10 \times 1 \times (283-273) = 80x + (50-x) \times 0.5 \times (273-263)$$
upon solving
$x= -2$g
which means that the ice doesnt melt instead it builds up, by freezing the water. Finally we have
$$ m_w=8g , m_{ice} = 52 g$$
now the entropy change :
$$ \Delta S_{Total} = \Delta S_{water}+\Delta S_{ice} $$
$$\Delta S_{water} = m_w C_w ln (\frac{T_f}{T_i})$$
$$\Delta S_{water} = 10 \times1\times ln (\frac{273}{283})$$
$$\Delta S_{ice} = \frac{-m_{ice} L}{T} + m_{ice} C_{ice} ln (\frac{T_f}{T_i})$$
$$\Delta S_{ice} = \frac{-2 \times 80}{273} + 50 \times 0.5 \times ln (\frac{273}{263})$$
$$\Delta S_{Total} = -0.00128 cal/K = -0.0535 J/K $$
the entropy change comes out to be negative but thats not possible, since its an irreversible process for an isolated system. According to the answer key
$$ \Delta S_{Total} = 0.1 J/K $$

Please find any conceptual or calculation error in my solution.

 

[mjc123]

the heat equation can be written as :
mwCwΔT=miceCiceΔT
10×1×(283−273)=80x+(50−x)×0.5×(273−263)

The equation is wrong because it does not include the latent heat. This is included in the calculation, but that is wrong because all the ice must warm up to 0C (and all the water cool to 0C).

 

[DaveC426913]

A 50 g cube of ice ... dropped in a container filled with 10 g water

Unless my calcs are wrong, that's an ice cube 3.68cm on a side dropped into a cubical container 3.91cm on a side, leaving 2.3mm all around (if you held it underwater).

³√50 = 3.68cm
³√(50+10) = 3.91cm

 

[Steve4Physics]

Since I don't know what the system will look like in the end I tried to check for the extreme cases.

Here’s a strategy to find the final state:

1) Calculate the amount of energy (A) needed to warm the ice to 0 deg.C.

2) Calculate the amount of energy (B) needed to melt the ice at 0 deg.C.

3) Calculate the amount of energy (C) lost by the water if it cools to 0 deg.C.

4) Calculate the amount of energy (D) lost by the (originally warm) water if it freezes at 0 deg.C. (You might want to leave this step till after step 5, as the value of D may not be needed.)

5) Use the values of A, B, C (and maybe D) to determine the final state. Then do any further calculations needed. For example, if C>A+B you know you will end up with all liquid at some final temperature, T>0. A further calculation is then needed to find the value of T.

 

[Su6had1p]

The equation is wrong because it does not include the latent heat. This is included in the calculation, but that is wrong because all the ice must warm up to 0C (and all the water cool to 0C).

Yes I realised that I missed the latent heat term in equation. But anyways included it in my calculations.
Also thanks for pointing out that the whole ice warms upto 0 deg C and not the 50-x thing. That was the source of my error. Thanks.