Question

Calorimetry Materials: Styrofoam cup & lid Digital thermometer Metal sample Graduated cylinder Beaker Test tube with stopper Heating setup for beaker Behind a calorimetry experiment is the known occurrence of heat flow from a warm object to a cooler object. Heat scales with the amount of matter and also depends on the resistance of an object to change temperature. Temperature is just the average of the amount of energy contained inside a given area. Heat is represented by the letter q, while temperature is measured by the letter T. This description is represented by the equation q-mCAT. The symbol "A" means change (or subtraction). The symbol C is the "resistance that was mentioned earlier, more specifically, this is the amount of energy that must be put into 1 gram of a given substance in order to raise the temperature by 1 °C. The specific heat of water is 4.184 J/g. "C. **When taking temperatures, do not allow the thermometer to touch any objects. ** Fill a 400mL beaker (or larger) three-fourths full with tap water. Begin heating the water; the water needs to boil before the experiment can begin. Find and record the mass of your Styrofoam cup. 0.93g Obtain 15 grams of the metal sample. Record the exact mass of your metal sample. 15g metal Once the water begins to boil, place your metal sample in a large test tube and gently cork the test tube. Place the test tube inside the boiling water. Leave the test tube in the boiling water for 10 minutes. This will heat up the metal sample, allowing this sample to be much hotter and higher in energy than the room temperature water that will be in the Styrofoam calorimeter Place 80mL of water inside your Styrofoam cup calorimeter, which is the water environment used for the heat transfer. Find the mass of the water that is inside your calorimeter. 80.93g - 0.93g = 80g HO Find the temperature of the water that is inside your calorimeter. 23°C

Answer 1

q lost & Agained zo -nmetal & Cmetal (Ts - Timetal) + "H20 TL 420 x (Tg - Tilko) = 0 15 (9) x Cmetal x (25.7- 100) + 8019) x 4.184 ( 19 ) (26.9-23) 20 - Cmetal x 15X(- J4.3) +903.944=o Cmetal = -903.744 - iliyos = 0.810896366 / gec = 0.81 5/9°c specific heat of metal = 0.81 / 8°C