Suppose we mix equal masses of water at different temperatures and measure the f
ID: 771901 • Letter: S
Question
Suppose we mix equal masses of water at different temperatures and measure the final temperature. In an ideal case, the final temperature should be the average of the two initial temperatures (since the masses of water samples are the same). We can calculate the energy involved as heat as follows: "heat lost" = (mass of higher temperature water)(cwater)(DeltaT) "heat gamed" = (mass of lower temperature water)(cwater)(DeltaT) where: DeltaT is defined as Tfinal - Tinitial cwater = heat capacity of water = 4.184 J/gdegreeC Ideally, the magnitude of "heat gained" by the cool water will be equal to the magnitude of the "heat lost" by the warm water (the signs will be different). In actuality, heat will be transferred to or from the calorimeter and the surroundings, resulting in a temperature that is different from the ideal case. In Part 1 of the lab you will test how well a coffee cup calorimeter matches the ideal case. You will do this by adding a known mass of warmer than room temperature water (at a known temperature) to a known mass of water at room temperature in the calorimeter. The relevant equations are: "heat lost" = (mhot water)(cwater)(DeltaThot water) "heat gained" = (mroom temp water)(cwater)(DeltaTroom temp water) + qcalorimeter where: qcalorimeter = (calorimeter constant)(DeltaTcalorimeter) qcalorimeter = DeltaTroom temp water As long as we use the same calorimeter, we do not need to know the mass of the calorimeter (the calorimeter constant is the product of the mass of the calorimeter and the heat capacity of the calorimeter). Ideally, the calorimeter constant would have a value of 0 J/degreeC. Consider the following data and answer the questions below. When 50.0 mL of 60.0degreeC water was mixed in the calorimeter with 50.0 mL of 25.0degreeC water, the final temperature was measured as 40.6 degreeC. Assume the density for water is 1.000 g/mL regardless of temperature. Determine the magnitude of the "heat lost" by the hot water. Do not enter a sign. Determine the magnitude of the "heat gained" by the room temperature water. Do not enter a sign. Determine the "heat gained" by the calorimeter. Be careful of the sign. Determine the calorimeter constant. You are given 6.7 grams of an unknown salt. You dissolve it in 55.5 mL of water. As the salt dissolves, the temperature of the solution changes from an initial temperature of 28.3degreeC to a final temperature of 54.7degreeC. Based on the chart of DeltaH values and the temperature change, you can determine which salt you have. What is the temperature change? This is the DeltaT component of the expression: qsurr = C times mass times DeltaT where qsurr is the heat lost or gained by the contents of the calorimeter due to the chemical reaction. What is the mass that is changing temperature in this experiment? Assume that the heat capacity, C, of the solution is the same as that of water, C = 4.184 J/g degreeC. Solve for qsurr. What is the value of qsys? The change in enthalpy, DeltaH. of this reaction is equal to qsys, since it was carried out under conditions of constant external pressure. Based on the sign of DeltaH (or qsys) the dissolution of the unknown salt is . When 6.7 g of salt are dissolved in 55.5 mL of water, 6880.4 J of heat are given off by the system. How much heat would be given off if only 1.00 g of salt had been dissolved in the water? In the table, the DeltaH values are given in kJ/mol. Convert your last answer to kJ to give kJ/g. Finally, multiply the kJ/g by the molar masses (g/mol) of possible salts, to get kJ/mol. The reaction of your unknown salt with water was exothermic. So, you will only need to look at salts with negative DeltaH values. formula of unknown = In order to determine the identity of your unknown salt you will determine DeltaHsoln in terms of J/g. The values of DeltaHsoln are generally given in terms of kJ/mol, so you will need to convert these. Complete the table below with values of DeltaHsoln in units of J/g.Explanation / Answer
A)-61170.08 J B)- 60317.4 J C)-852.68 D)- 54.66