based on this experiment, how do I obtain #2 heat gsined by water (J) and #4 spe

Question

based on this experiment, how do I obtain #2 heat gsined by water (J)
and #4 specific heat of metal, equstion
please explain

Calorimetry Desk No. Date A. Specific Heat of a Metal Unknown No, Lab Sec. Name Trial I Trial 2 0 1, Mass of metal (g) 2. Temperature of metal (boiling water) cc) 3. Mass of calorimeter (g) 4. Mass of calorimeter + water (g) 5. Mass of water (g) 6. Temperature of water in calorimeter (°C) 7 Maximum temperature of metal and water from graph (°C n·2 8. Instructor's approval of graph Calculations for Specific Heat of a Metal 1. Temperature change of water, T("O 2. Heat gained by water (J) 3, Temperature change of metal, TC 4. Specific heat of metal, equation 255 (Me-c) 5. Average specific heat of metal (Mg-c) Show calculations for Trial 1 using the correct number of significant figures. o e 41C26. Data Analysis, B Experiment 25 301 131

Explanation / Answer

Ans. Amount of heat changed (gained or lost) during attaining thermal equilibrium is given by-

            q = m s dT                            - equation 1

Where,

q = heat gained or lost

m = mass

s = specific heat

dT = Final temperature – Initial temperature

# At thermal equilibrium, total heat lost by hot metal (immersed in water bath) is equal to the amount of heat gained by water in calorimeter. To depict heat loss by meat, a –ve sign is used.

# Trial 1:

Mass of water in calorimeter = 74.514 g – 5.587 g = 68.927 g

Initial temperature of metal = 99.70C

Initial temperature of calorimeter = 25.00C

Temperature at thermal equilibrium = 31.40C

Therefore, final temperature in calorimeter = 31.40C

Now,

            - q1 (metal) = q2 (water)

            Or, - 28.715 g x s1 x (31.4 – 99.7)0C = 68.927 g x (4.184 J g-10C-1) x (31.4 – 25.0)0C

            Or, - 28.715 g x s1 x (-68.3)0C = 68.927 g x (4.184 J g-10C-1) x (6.4)0C

            Or, 1961.2345 g 0C x s1 = 1845.6996352 J

            Or, s1 = 1845.6996352 J / (1961.2345 g 0C)

            Hence, s1 = 0.9411 J g-10C-1

Therefore, specific heat of metal, s1 = 0.9411 J g-10C-1

# Trial 2:

Mass of water in calorimeter = 74.383 g – 5.601 g = 68.782 g

Initial temperature of metal = 100.40C

Initial temperature of calorimeter = 23.60C

Temperature at thermal equilibrium = 33.70C

Therefore, final temperature in calorimeter = 33.70C

Now,

            - q1 (metal) = q2 (water)

            Or, - 29.008 g x s1 x (33.7 – 100.4)0C = 68.782 g x (4.184 J g-10C-1) x (33.7 – 23.6)0C

            Or, - 29.008 g x s1 x (-66.7)0C = 68.782 g x (4.184 J g-10C-1) x (10.1)0C

            Or, 1934.8336 g 0C x s1 = 2906.6172688 J

            Or, s1 = 2906.6172688 J / (1934.8336 g 0C)

            Hence, s1 = 1.50 J g-10C-1

Therefore, specific heat of metal, s1 = 1.50 J g-10C-1

Note: The specific heat of meat varies a lot in the two trials. Please recheck the input values recorded during experiment.

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