Question

Calorimetry Lab
-Finding speficic heat capacity of calorimeter C(cal)

Exp.5 CHEMICAL EQUILIBRI 2. (2 marks) Table 1: Determining the heat capacity of the calorimeter. Run 1 Run 2 Run 3 Mass of the hot water, m. (g): Initial temperature of the hot water, T. (°C): Mass of the cold water, me (g): Initial temperature of the cold water and calorimeter, T. (°C): Final temperature of the mixture, T. (°C): Heat lost by the hot water, Qh (J): Heat gained by the cold water, Qc (J): Heat gained by the calorimeter, Qcal (J): Heat capacity of the calorimeter, C (Jºc): Average heat capacity of calorimeter, Ceal (J °C): 3. (1 mark) In this space, please show the full set of calculations for Part 1 Run 1.

I need this by today pls help!

Answer 1

Answer: Weight of hot water= 15 g

temperature of hot water= 86oC

Temperature of the mixture= 35.4oC

Temperature difference = 50.6 oC

Specific Heat capacity of water Cp= 4.184 J/g oC

Heat lost = q1= m1 $\large \Delta$ T Cp = 15x50.6x 4.184 = 3175.6 J

Weight of cold water= 15 g

temperature of hot water= 23.4oC

Temperature of the mixture= 35.4oC

Temperature difference = 12 oC

Specific Heat capacity of water Cp= 4.184 J/g oC

Heat gained = q2= m2 $\large \Delta$ T Cp = 15x12x 4.184 = 753.12J

The heat gained by the calorimeter= 3175.6-753.12 = 2422.48 J

Heat capacity of the calorimeter= 2422.48/12 = 201.87 J/oC.