\"heat lost\" = (mass of higher temperature water)(cwater)(T) \"heat gained\" =

Question

"heat lost" = (mass of higher temperature water)(cwater)(T) "heat gained" = (mass of lower temperature water)(cwater)(T) where: T is defined as Tfinal – Tinitial cwater = heat capacity of water = 4.184 J/g°C 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)(Thot water) “heat gained” = (mroom temp water)(cwater)(Troom temp water) + qcalorimeter where: qcalorimeter = (calorimeter constant)(Tcalorimeter) Tcalorimeter = Troom 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/°C. Consider the following data and answer the questions below. When 50.0 mL of 60.0°C water was mixed in the calorimeter with 50.0 mL of 25.0°C water, the final temperature was measured as 41.3 °C. Assume the density for water is 1.000 g/mL regardless of temperature. Determine the magnitude of the heat lost by the hot water.

Explanation / Answer

From the given data,

mass of hot water = 50 ml x 1 g/ml = 50 g

Cp = 4.184 J/g.oC

dT = (60 - 41.3) = 18.7 oC

heat lost by hot water (q1) = mCpdT

                                           = 50 x 4.184 x 18.7

                                           = 3912.04 J

mass of cold water = 50 ml x 1 g/ml = 50 g

Cp = 4.184 J/g.oC

dT = (41.3 - 25) = 16.3 oC

heat lost by cold water (q2) = mCpdT

                                            = 50 x 4.184 x 16.3

                                            = 3409.96 J

heat gained by calorimeter = 3912.04 - 3409.96

                                           = 502.08 J

Heat capacity of calorimeter = 502.08 J/16.3 oC

                                              = 30.802 J/oC

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