Question

Review | Constants Periodic Table Part A Assume you are performing the calibration step of Experiment 8 and you begin with 30 g of water at 20 °C and 30 g of water at 80 °C. After adding the two portions of water into your calorimeter setup and following the procedure outlined in the experiment, you determine the temperature of the mixed portions of water to be 45 °C. What is the heat capacity of the calorimeter? Assume room temperature is 25 °C. IVO AD * O ? calc Submit Request Answer Provide Feedback Next >

Answer 1

Mass of cool water, m = 30 g heat capacity of water, C, = 4.184J/gºC initial temperature of water, 7= 20 °C initial temperature of water, T, = 45°C Heat transferred to cool water, 9coot = mc, (TE-T) = 30 gx4.184 JigºCx(45 °C–20 °C) = 3138J --- ------ -------- ------ ------ Mass of hot water, m = 30 g heat capacity of water, C, = 4.184J/g°C initi al temperature of water, 7. = 80 °C initial temperature of water, T, = 45°C Heat transferred from hot water, “hot = ,C, T-6) = 30 gx4.184J/g°Cx(45 °C-80 °C) = -4393.2 J - - - - - - - - - - - - - - - - - - - - - - - - Let Cray be the heat capacity of calorimeter Initi al temperature of calorimeter, 7. = 25°C Initi al temperature of calorimeter, T, = 45°C Heat absorbed by the calorimeter is 9cas = Co(T-) = Coal (45°C-25°C) = 20CC - - - - - - - - - - - - - - - - - - - - - - - - - Write the heat balance as follows: cool + cal = - Chat Substitute the corresponding values in the above equation and calculate Ceat 3138J+20C =-(-4393.2 J) 4393.2 J-3138J C 20 °C Coa = 62.76J/°C Therefore, the required solution is 62.76J/°C