.................................................................Trial A........

ID: 1046380 • Letter: #

Question

.................................................................Trial A.................. Trial B

Mass of metal: ____________15.039 g._______6.847 g

Mass of water: _____________55.93 g_______59.657 g

Initial temp of metal:_________95.2 C_________ 95 C

Initial temp of water:__________23.3 C________23.8 C

FInal temp of water and metal: __24.2 C_______24.3 C

Change in temp of metal: _______71 C________70.4 C

Change in temp of water:______ .8 C_________ .4 C

Please Determine:

a. Heat flow of water (include correct sign)

b. Heat flow of metal (include correct sign)

c. Specific heat of metal

d. Average specific heat of metal

e. Approximate molar mass and element symbol of metal

Please show all calculations. I am not sure where to even start. Thank you in advance.

Explanation / Answer

Trail A

Trial B

Heat flow of water = (mass of water)*(specific heat capacity of water)*(change in temperature of water)

[Specific heat capacity of water = 4.184 J/g.°C]

(55.930 g)*(4.184 J/g.°C)*(0.8°C) = 187.209 J

(59.657 g)*(4.184 J/g.°C)*(0.4°C) = 99.842 J

Heat flow of metal

The heat flow of the metal is equal in magnitude, but opposite in sign to the heat flow of water. Hence, the heat flow of the metal will be the negative of the heat flow of water.

-187.209 J

-99.842 J

Specific heat of metal

Let the specific heat of the metal be S. Thus, the heat flow of the metal = (mass of metal)*(specific heat capacity of the metal)*(final temperature of the metal – initial temperature of the metal).

(15.039 g)*S*(71 – 95.2)°C = -187.209 J

====> -363.9438*S g.°C = -187.209 J

====> S = (-187.209 J)/(-363.9438 g.°C) = 0.514 J/g.°C

(6.847 g)*S*(70.4 – 95.0)°C = -99.842 J

====> -168.4362*S g.°C = --99.842 J

====> S = (-99.842 J)/(-168.4362 g.°C) = 0.593 J/g.°C

Average specific heat of metal

½*(0.514 + 0.593) J/g.°C = 0.5535 J/g.°C

Molar mass of metal

As per Dulong-Petit’s law,

Molar mass of metal = 25 J/mol.°C/(specific heat of metal)

====> Molar mass of metal = (25 J/mol.°C)/(0.5535 J/g.°C)

====> Molar mass of metal = 45.167 g/mol