A styrofoam cup calorimeter is used to determine the heat ofreaction for the dis

Question

A styrofoam cup calorimeter is used to determine the heat ofreaction for the dissolution of an ionic compound in water. When4.50 g of this compound is added to 300.0 mL of water, thetemperature rises from 18.3°C to 25.8°C. Calculate the heatof reaction (qrxn = Hrxn) per gram ofcompound (in kJ/g).
For dilute aqueous solutions, the specific heat isapproximately the same as that of water, 4.18J/(g·°C).
For dilute aqueous solutions, the specific heat isapproximately the same as that of water, 4.18J/(g·°C).

Explanation / Answer

         Pockets of air inside the Styrofoam providesbetter insulating qualities than paper. Hence a Styrofoam cup makesa better calorimeter than a paper cup. The given data is    Mass of the compound m= 4.50 g    The initial temperature t1=18.30 C    The final temperature   t2 =25.80 C    The change in temperature t=t2 - t1                                            =    25.80C -18.30C                                              = 7.50C Specific Heat of theWater s =  4.186joule/gram °C Heat of the reaction qrxn =m*s*t                                          = 4.50 gram *4.186 joule/gram °C * 7.50C                                           = 141.2775 Joule /gram                                          = 141.2775 *10-3 kJ/gram                                            =0.1412775 kJ/gram     The Heat of the reaction is0.1412775 kJ/gram    Pockets of air inside the Styrofoam providesbetter insulating qualities than paper. Hence a Styrofoam cup makesa better calorimeter than a paper cup. The given data is    Mass of the compound m= 4.50 g    The initial temperature t1=18.30 C    The final temperature   t2 =25.80 C    The change in temperature t=t2 - t1                                            =    25.80C -18.30C                                              = 7.50C Specific Heat of theWater s =  4.186joule/gram °C Heat of the reaction qrxn =m*s*t                                          = 4.50 gram *4.186 joule/gram °C * 7.50C                                           = 141.2775 Joule /gram                                          = 141.2775 *10-3 kJ/gram                                            =0.1412775 kJ/gram     The Heat of the reaction is0.1412775 kJ/gram

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