Calculate the specific heat (s) of the metal for trial 1: Calculate the specific

Question

Calculate the specific heat (s) of the metal for trial 1: Calculate the specific heat (s) of the metal for trial 2: Specific heat: Trial 1 Trial 2 Average Find the density for the metal from your report for Experiment 2 (part B). Check the table of densities and specific heats (posted in lab and on Moodle) to identify the one metal that best agrees with your specific heat and density. Density of the metal (expt 2B): Best reasonable GUESS as to the identity of the metal: A calorimeter contains 180.0 mL of water at 5.0degreeC. What is T_final f a 55.00 g sample of your metal at 305.0degreeC is added? Use the average specific heat you measured. If a sample of your metal at 220.0degreeC causes a 120.0 mL sample of water to increase in temperature from 21.0degreeC to 24.3degreeC, what is mass of the metal? Use the average specific heat you measured. How much heat must be released to cool a 0.18 mol sample of S (s= 0 71 j/g^degree C) from 18.0 to -24.5degreeC?

Explanation / Answer

Ans.     q = m c x dT     - equation 1

            where, q = heat change

                        m = mass

c = specific heat of the object

dT = final – initial temperature (if heat is being gained) ;

            or, (Initial- final temperature) if heat is being lost

Note: the +ve or -ve value of dT just indicates the heat gain or loss, respectively, without changing the quantity of heat.

Calculate:

Trial 1: dT for metal = 95.00C – 25.40C = 69.60C

            dT for water = 25.40C – 24.20C = 1.20C   

Trial 2: dT for metal = 95.00C – 24.80C = 70.20C

            dT for water = 24.80C – 23.20C = 1.60C

Calculate specific heat in Trial 1: Heat lost be metal is equal to heat gained by water. So,

            q metal = q water

or, m1 c1 x dT1 = m2 c2 x dT2             ; [1=for metal , 2 = for water]

or, 22.81 g x c1 x (69.60C) = 84.90 g x (4.18 J/g0C) x (1.20C)           ; [specific heat of water = 4.18 J/g0C ]

or, 1587.576 g0C x c1 = 425.8584 J

or, c1 = 425.8584 J / (1587.576 g0C) = 0.2682 J/g0C

Thus, specific heat of metal, trial 1 = 0.2682 J/g0C

Calculate specific heat in Trial 2:

            22.81 g x c1 x (70.20C) = 85.05 g x (4.18 J/g0C) x (1.60C)

            Or, 1601.262 g0C x c1 = 568.8144 J

            Or, c1 = 568.8144 J / (1601.262 g0C) = 0.3552 J/g0C

Thus, specific heat of metal, trial 2 = 0.3552 J/g0C

Average specific heat of metal = average of specific heat in trial 1 and 2

                                                = (0.2682 J/g0C + 0.3552 J/g0C) / 2

                                                = 0.3117 J/g0C

Ans. a. Let the final temperature be T.

Heat lost be metal = heat gained by water

q metal = q water

or, m1 c1 x dT1 = m2 c2 x dT2             ; [1=for metal , 2 = for water]

or, 55.00 g x (0.3117 J/g0C) x (305.00C - T) = 180.00 g x (4.18 J/g0C) x (T- 50C) ;[density (water = 1.0 g/ml]

or, 17.4735 J/0C x (305.00C - T) = 752.4 J/0C x (T- 50C)

or, 5329.4175 J – (17.4735 J/0C) T = (752.4 J0C) T - 3762 J

or, 5329.4175 J + 3762 J = (752.4 J0C) T + (17.4735 J/0C) T

or, T = 9091.4175 J / (769.8735 J/0C) = 11.810C

Thus, final temperature = 11.810C

Ans. b. Heat gained by water, q = 120.0 g x (4.18 J/g0C) x (24.3 – 21.0)0C

Or, q = 1655.28 J

Using, heat gained by water = heat lost by metal

So, 1655.28 J = q (metal) = m x (0.3117 J/g0C) x (2200C – 24.30C)

Or, m = 1655.28 J / (60.99969 J/g) = 27.14 g

Thus, mass of metal = 27.14 g

Ans. c. In absence of identity of the sample and/or data relevant for this, the calculation can’t be made using “moles” of sample S.

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