Ice calorimetry lab questions

  • This question is about the lab we did in class: Ice Calorimetry Lab.
    1. Plot your data I Plot the temperature versus total energy added to the system (which you can call \(Q\)). To do this, you will need to integrate the power. Discuss this curve and any interesting features you notice on it.
    2. Plot your data II Plot the heat capacity versus temperature. This will be a bit trickier. You can find the heat capacity from the previous plot by looking at the slope. \begin{align} C_p &= \left(\frac{\partial Q}{\partial T}\right)_p \end{align} This is what is called the heat capacity, which is the amount of energy needed to change the temperature by a given amount. The \(p\) subscript means that your measurement was made at constant pressure. This heat capacity is actually the total heat capacity of everything you put in the calorimeter, which includes the resistor and thermometer.
    3. Specific heat From your plot of \(C_p(T)\), work out the heat capacity per unit mass of water. You may assume the effect of the resistor and thermometer are negligible. How does your answer compare with the prediction of the Dulong-Petit law?
    4. Latent heat of fusion What did the temperature do while the ice was melting? How much energy was required to melt the ice in your calorimeter? How much energy was required per unit mass? per molecule?
    5. Entropy of fusion The change in entropy is easy to measure for a reversible isothermal process (such as the slow melting of ice), it is just \begin{align} \Delta S &= \frac{Q}{T} \end{align} where \(Q\) is the energy thermally added to the system and \(T\) is the temperature in Kelvin. What is was change in the entropy of the ice you melted? What was the change in entropy per molecule? What was the change in entropy per molecule divided by Boltzmann's constant?
    6. Entropy for a temperature change Choose two temperatures that your water reached (after the ice melted), and find the change in the entropy of your water. This change is given by \begin{align} \Delta S &= \int \frac{{\mathit{\unicode{273}}} Q}{T} \\ &= \int_{t_i}^{t_f} \frac{P(t)}{T(t)}dt \end{align} where \(P(t)\) is the heater power as a function of time and \(T(t)\) is the temperature, also as a function of time.