Lab Instructor EXPERIMENT 5 Specific Heat Capacity Results/Observations A. Speci

Question

Lab Instructor EXPERIMENT 5 Specific Heat Capacity Results/Observations A. Specific Heat of a Known Metal 1. Mass of the coffee-cups L 105g -, Mass of the cups+ water 113204 Mass of the water Temprature of col water 2. Identity of metal sample 20. 5 Zinc Mass of metal sample Temperature of boiling water Maximum System Temperature (T)22 ple 3.lo 1.2 t.o 21. 3. 4 Calculate the specific heat of the known and percent error. Show work! C,= J/g· % error- 0 2012 Cmpage Learning All Righs Reservod. May sot be scansed, copied or 49 Righs Reserved. May sot be wassed, copied or daplicateod, er posod so a publicly accessibile website,in whole or in pan

Explanation / Answer

Ans. #1. 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.

So,

            - q1 (metal) = q2 (water)

            Or, - 39.116 g x s1 x (22.4 – 97.2)0C = 113.204 g x (4.184 J g-10C-1) x (22.4 – 20.9)0C

            Or, 2925.8768 s1 g 0C= 710.46830 J

            Or, s1 = 710.46830 J / (2925.8768 g 0C) = 0.2428 J g-10C-1

Therefore, specific heat of Zn, s1 = C1 = 0.2428 J g-10C-1

# Actual specific heat of Zn = 0.387 J g-10C-1

Error = Actual value – experimental value = 0.387 J g-10C-1 - 0.2428 J g-10C-1

= 0.1442 J g-10C-1

Now,

            % Error = (Error / Actual value) x 100

                                    = (0.1442 J g-10C-1 / 0.387 J g-10C-1) x 100

                                    = 37.26 %

#2. 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.

So,

            - q1 (metal) = q2 (water)

            Or, - 39.116 g x s1 x (22.7 – 97.2)0C = 154.577 g x (4.184 J g-10C-1) x (22.7 – 21.1)0C

            Or, 2914.142 s1 g 0C= 757.8329 J

            Or, s1 = 757.8329 J / (2914.142 g 0C) = 0.260 J g-10C-1

Therefore, specific heat of Zn, s1 = C1 = 0.260 J g-10C-1

# Actual specific heat of Zn = 0.387 J g-10C-1

Error = Actual value – experimental value = 0.387 J g-10C-1 - 0.260 J g-10C-1

= 0.127 J g-10C-1

Now,

            % Error = (Error / Actual value) x 100

                                    = (0.127 J g-10C-1 / 0.387 J g-10C-1) x 100

                                    = 32.82 %

# Average specific heat = (0.2428 J g-10C-1 + 0.260 J g-10C-1) / 2 = 0.2514 J g-10C-1

# Average % error = (37.26 % + 32.82 %) / 2 = 35.04 %

#B. 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.

So,

            - q1 (metal) = q2 (water)

            Or, - 23.430 g x s1 x (21.7 – 96.8)0C = 65.087 g x (4.184 J g-10C-1) x (21.7 – 20.7)0C

            Or, 1759.593 s1 g 0C= 272.324 J

            Or, s1 = 272.324 J / (1759.593 g 0C) = 0.155 J g-10C-1

Therefore, specific heat of the unknown 2, s1 = C1 = 0.155 J g-10C-1

# Identity of the unknown: Lutetium, s = 0.150 J g-10C-1

Please use you reference book to point out any other element.

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