Energy and Entropy: Winter-2026
Lab 3: Melting Ice (SOLUTION): Due Day 17 W6 D3

1. Melting ice lab questions S1 5458S These questions relate to the in-class activity where you put ice (prepared at zero degree celcius) into some warm water. You insulated the ice and water inside nested polystyrene cups and closed the lid. You waited a few minutes for the system to reach equilibrium. At equilibruium, all the melted ice and water were at the same temperature. During class, you started to calculate a prediction for this final temperature.
   1. Prediction What is your prediction for the final temperature? List the measured quantities that you used in your calcultion. Show your work.
   2. Measurement What final temperature did you measure? Comment on the magnitude and sources of errors in your experiment and in your prediction.
   3. Change in entropy Calculate the total change in entropy that occured during the overall process. Initial state: warm water and ice. Final state: cool water at a uniform temperature.

      If warm water cools down from 313 K to 283 K, in a quasi-static process, the entropy change is \begin{align} \Delta S_{water} &= \int \frac{{\mathit{\unicode{273}}} Q}{T}\\ &= \int_{T_i}^{T_f} \frac{C_pdT}{T}\\ &= C_p \ln\left(\frac{T_f}{T_i}\right)\\ &= C_p \ln\left(\frac{283\text{ K}}{313\text{ K}}\right)\\ &= -0.1008C_p\\ & \Delta S_{\text{water}} \approx-117 \frac{J}{K} \end{align} where \(C_p = M_{\text{water}}c_p\) is the heat capacity, where \(c_p=4.183 \frac{J}{gK}\), the specific heat of water. The actual water had this initial temperature and final temperature, so the actual change in entropy must be the same as the quasi-static process that I just calculated.

      Now, I'll imagine melting the ice in a series of two quasi-static processes: (1) the phase change, (2) warming the water that was formerly ice. For the phase change, imagine keeping temperature at 273 K throughout the melt process. This gives a change in entropy \begin{align} \Delta S_{\text{ice,step1}} &= \int \frac{{\mathit{\unicode{273}}} Q}{T}\\ &= \frac{l_f M_{\text{melted}}}{T}\\ &= \frac{(333 \frac{J}{g})(103g)}{273\text{ K}}\\ &= \Delta S_{\text{ice}} \approx 126 \frac{J}{K} \end{align}

      Now, warming up the melted ice from 273 K to 283 K, in a quasi-static process, the entropy change would be \begin{align} \Delta S_{\text{ice,step2}} &= \int \frac{{\mathit{\unicode{273}}} Q}{T}\\ &= \int_{T_i}^{T_f} \frac{C_pdT}{T}\\ &= C_p \ln\left(\frac{T_f}{T_i}\right)\\ &= C_p \ln\left(\frac{313\text{K}}{273\text{K}}\right)\\ &= +0.1367C_p\\ & \Delta S_{\text{ice,step2}} \approx+117 \frac{J}{K} \end{align}

      Now, we can add up all the entropies: \begin{align} \Delta S_{net}&=\Delta S_{\text{water}}+\Delta S_{\text{ice,step1}} +\Delta S_{\text{ice,step2}} \\ &=(126 \frac{J}{K})+(-117 \frac{J}{K})+ something \\ &=9 \frac{J}{K} \end{align} So the entropy of our system increased! This is good, because our process was not quasistatic or reversible,so entropy shouldn't be 0 for the process. The 2nd Law tells us it can only go up for an isolated system that has done an irreversible process.