Date READ THE EXPERIMENTAL DISCUSSION FIRST! SHOW CALCULATION SET-UP on the next

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Date READ THE EXPERIMENTAL DISCUSSION FIRST! SHOW CALCULATION SET-UP on the next page. Report ALL answers to correct significant figures PART A:Specific Heat of an Unknown Metal Mass of Styrofoam calorimeter cup Mass of Styrofoam calorimeter cup and water (40-45 mL water) Initial water temperature Final temperature of metal/water mixture Mass of Styrofoam calorimeter cup, water, and metal Initial metal temperature (Boiling water bath temperature) 5.676 g 44276 g 25.2 C 32.5? 89.476 g 100.0°C CALCULATIONSUMMARY Specific heat of water Temperature change of water Mass of water Temperature change of metal Mass of metal Specific Heat of a Metal (Show "set-up"below) 4.184J/g·? J/g·? mole Approximate Gram Atomic Weight

Explanation / Answer

Ans. Part A:

# Temperature change of water, dT1 = (final – initial) temp. = 32.50C – 25.20C = 7.30C

# Mass of water = 44.276 g – 5.676 g = 38.60 g

# Temperature change of metal, dT2 = 32.50C – 100.00C = -67.50C

# Mass of metal = 89.476 g – 44.276 g = 45.20 g

# When hot metal is placed in contact with cold water, the hot metal loses some heat, and the water gains the same amount of heat in order to attain the thermal equilibrium.

Amount of heat gained/lost by the samples is given by-

q = m s dT                            - equation 1

Where,

q = heat gained/lost

m = mass of sample

s1 = specific heat of sample

dT = Final temperature – Initial temperature

Now,

            Heat lost by metal = Heat gained by water

            Or, -q (metal) = +q (water)             ; the –ve sign indicates heat loss

            Or, -[45.20 g x s x (-67.5)0C] = 38.60 g x 4.184 J g-10C-1 (7.3)0C

            Or, 3051.0 s g 0C = 1178.96752 J

            Or, s = 1178.96752 J / 3051.0 s g 0C = 0.386 J g-10C-1

Therefore, specific heat of the unknown meatal = 0.386 J g-10C-1

Part B: Mass of solution = 57.242 g – 5.742 g = 51.50 g

# Temperature change for solution, dTsoln. = 37.10C – 24.00C = 13.100C

# Mass of solid sample = 8.242 g – 5.742 g = 2.50 g

# Moles of solid sample = Mass / MW = 2.50 g / 40.0 g mol-1 = 0.0625 mol

# Heat flow (q) for the above sample size:

Amount of heat gained by the solution while increase in its temperature is given by-

q = m s dT                            - equation 1

Where,

q = heat gained

m = mass of sample

s1 = specific heat of sample

dT = Final temperature – Initial temperature

Putting the values in equation 1-

            q = 51.50 g x 4.184 J g-10C-1 x 13.100C = + 2822.7356 J

# Enthalpy change for above sample size:

The total heat gained by solution must be equal to the amount of heat released during dissolution of solid sample.

So,

            dH = -q = - 2822.7356 J

# Enthalpy change for 1 mol solid = dH / moles of solid

                                                            = (-2822.7356 J) / 0.0625 mol

                                                            = -45163.7696 J/mol

                                                            = -45.164 kJ /mol

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