Experiment 9 Report Sheet Scali Data: I UNITS!!! Metal Sample 1: E Tria12. A Mas

Question

Experiment 9 Report Sheet Scali Data: I UNITS!!! Metal Sample 1: E Tria12. A Mass of metal sample B Mass of empty calorimeter 1946 C Mass of calorimeter plus water 33.08 D Initial temperature of calorimeter water ao C 80.0 C Initial temperature of metal boiling temp of H20) LDO Equilibrium temperature of calorimeter water and metal (final temperature) NCLUDE UNITS AND SHOW WORK (attach or use back of pagentit scratch er G Mass of calorimeter water H ATwater AT K qmetal L Specific heat of Metal Sample 1 M Average specific heat of Metal Sample 1 mpare your average specific heat for Metal Sample 1 to the value in Table 9-1. Calculate e percent error in your result. 9.6

Explanation / Answer

Given that

For calorimeter, we know that, the heat lost by metal block is equal to heat gained by calorimeter water.

Hence for trial 1,

Mass of calorimeter water = Mass of calorimeter plus water - mass of empty calorimeter

Mass of calorimeter water = [C] - [B]

Mass of calorimeter water = 73.08 - 17.09

Mass of calorimeter water = 55.99 gms = [G]

Change in water temperature = Final temperature of water - initial temperature of water

Change in water temperature = equillibrium temperature of water and metal - initial temperature of water

Change in water temperature = [F] - [D]

Change in water temperature = 22.6 - 20

Change in water temperature = 2.600C = [H]

we know, for water, specific heat [Cp] is 4.184 J/g.°C

Heat lost by water, qwater = mwater*Cpwater*Twater

qwater = 55.99 * 4.184 * 2.6

qwater = 609.08 J = [I]

Change in metal temperature = Initial metal temperature - final metal temperature

Change in metal temperature = Initial metal temperature - equillibrium metal temperature

Change in metal temperature = [E] - [F]

Change in metal temperature = 100 - 22.6

Change in metal temperature = 77.40C = [J]

we know, the heat lost by metal block is equal to heat gained by calorimeter water.

hence qwater = qmetal = 609.08 J = [K]

Heat gained by metal block, qmetal = mmetal*Cpmetal*Tmetal

609.08 J = 26.37 *Cpmetal*77.4

Cpmetal = 0.298 J/g.°C = [L]

For trial 2

Mass of calorimeter water = Mass of calorimeter plus water - mass of empty calorimeter

Mass of calorimeter water = [C] - [B]

Mass of calorimeter water = 70.81 - 19.48

Mass of calorimeter water = 51.33 gms = [G]

Change in water temperature = Final temperature of water - initial temperature of water

Change in water temperature = equillibrium temperature of water and metal - initial temperature of water

Change in water temperature = [F] - [D]

Change in water temperature = 22.2 - 20

Change in water temperature = 2.20C = [H]

we know, for water, specific heat [Cp] is 4.184 J/g.°C

Heat lost by water, qwater = mwater*Cpwater*Twater

qwater = 51.33 * 4.184 * 2.2

qwater = 472.48 J = [I]

Change in metal temperature = Initial metal temperature - final metal temperature

Change in metal temperature = Initial metal temperature - equillibrium metal temperature

Change in metal temperature = [E] - [F]

Change in metal temperature = 100 - 22.2

Change in metal temperature = 77.80C = [J]

we know, the heat lost by metal block is equal to heat gained by calorimeter water.

hence qwater = qmetal = 472.48 J = [K]

Heat gained by metal block, qmetal = mmetal*Cpmetal*Tmetal

472.48 J = 31.74 *Cpmetal*77.8

Cpmetal = 0.191 J/g.°C = [L]

Average specific heat of metal = (0.191 + 0.298)/2 = 0.2445 J/g.°C = [M]

Trial 1 Trial 2 A Mass of metal sample 27.37 31.74 B Mass of empty calorimeter 17.09 19.48 C Mass of calorimeter + water 73.08 70.81 D Initial temp of calorimeter water 200C 200C E Initial temp of metal 1000C 1000C F Equillibrium temp of calorimeter water and metal 22.60C 22.20C

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