I\'m trying to do the data analysis for a calorimetry lab and need help answerin
ID: 816526 • Letter: I
Question
I'm trying to do the data analysis for a calorimetry lab and need help answering this question using my data. Can someone please explain how to solve this for the first two trials so I can finish the rest?
A 139.29g solid aluminum cylinder was heated in a boiling bath (100 degrees C) of 200 mL of water for 5 minutes before being immersed in room temperature water in a constant volume calorimeter (a coffe cup calorimeter). Calculate the heat gained by the water in each of the trials and use this to determine the heat lost by the aluminum bar and then calculate an experimental value for the specific heat of aluminum using the equation below:
Information needed to answer the question:
The specific heat capacity of water is 4.184J/g degree Celsius
Mass of water in trial 1 is 40.629g. Initial water temperature is 20.0 degrees Celsius. Final water temperature after aluminum cylinder immersion is 25.2 degrees Celsius.
Mass of water in trial 2 is 45.153g. Initial water temperature is 21.1 degrees Celsius. Final water temperature after aluminum cylinder immersion is 25.3 degrees Celsius.
I'm trying to do the data analysis for a calorimetry lab and need help answering this question using my data. Can someone please explain how to solve this for the first two trials so I can finish the rest? A 139.29g solid aluminum cylinder was heated in a boiling bath (100 degrees C) of 200 mL of water for 5 minutes before being immersed in room temperature water in a constant volume calorimeter (a coffe cup calorimeter). Calculate the heat gained by the water in each of the trials and use this to determine the heat lost by the aluminum bar and then calculate an experimental value for the specific heat of aluminum using the equation below: q=mc delta T Information needed to answer the question: The specific heat capacity of water is 4.184J/g degree Celsius Mass of water in trial 1 is 40.629g. Initial water temperature is 20.0 degrees Celsius. Final water temperature after aluminum cylinder immersion is 25.2 degrees Celsius. Mass of water in trial 2 is 45.153g. Initial water temperature is 21.1 degrees Celsius. Final water temperature after aluminum cylinder immersion is 25.3 degrees Celsius.Explanation / Answer
Heat lost by Al bar = Heat gained by water
Trial 1 : Q= m x c x(delta T) = 40.629 g x 4.184 J/gC x 5.2 C = 883.96 J
Similarly for Trial 2 , Q = 45.153 g x 4.184 J/g C x 4.2 C = 793.46 J
Average Q = 838.71 J
Now apply this value back to Al,
c of Al = Q avg / (m x Delta T)
= 838.71 J / (139.29 g x 74.75 C)
= 0.081 Jg-1C-1