A 100.0 g sample of an unknown metal at 90.0 degree C is thrown into a calorimet

Question

A 100.0 g sample of an unknown metal at 90.0 degree C is thrown into a calorimeter containing 50.0 g of water at 25.0 degree C. After the system comes to equilibrium, the temperature is 31.6 degree C. If no heat is lost to the environment or the calorimeter, calculate the heat capacity of the metal. The heat capacity of water is 4.184 J/gK. 8. A gas occupies a volume of 0.600 l. at 36 degree C and 1.00 atm pressure. What will its volume be at 0 degree C and 150.0 mmHg? Calculate the mass density of a sample of argon gas at-15.0T and 100 atm in a 1.00 L flask.

Explanation / Answer

7) In this experiment the unknown metal liberates heat and water absorbs heat and we have to calculate them:

In water, Given:

m = 50.0 g
C = 4.184 J/gK=4.184 J/gC
Tinitial = 25.0°C
Tfinal = 31.6°C
T = 6.6°C (Tfinal - Tinitial)

Solve for Qwater:

Qwater = m•C•T = (50.0 g)•(4.184 J/g°C)•(6.6°C)
Qwater = 1,381 J
(The + sign indicates that heat is absorbed by the water)

For unknown metal, Given:

Qmetal = -1381  J (use a - sign since the metal is losing heat)
m = 100.0 g
Tinitial = 90.0°C
Tfinal = 31.6°C
T = (Tfinal - Tinitial ) = -58.4°C

Solve for Cmetal:

Rearrange Qmetal = mmetal•Cmetal•Tmetal to obtain Cmetal = Qmetal / (mmetal•Tmetal)

Cmetal = Qmetal / (mmetal•Tmetal) = (-1,381 J)/[(100.0g)•(-58.4°C)]
Cmetal = 0.2365 J/g°C
Cmetal = 0.237 J/g°C (rounded to three significant digits)

8) This is a gas problem, using combined law:

V1P1T2 = V2P2T1

P is pressure, T is temperature in K and V is volume. Number 1 indicates initial and number 2 is final.

Data given:

V1 = 0.600L

P1 = 1atm

T1 = 36°C.........   36 + 273 = 309K

V2 = ?

P2 = 150.0mmHg.....   150.0mmHg(760.0 mmHg/1 atm) = 0.1974 atm

T2 = =0°C   = 273K

Lets calculate V:

In equation   V1P1T2 = V2P2T1   you have that:

V1P1T2/ P2T1 = V2

V2 =V1P1T2/ P2T1 = (0.600L)(1.00atm)(273K) / (0.1974 atm)(309K) = 2.69L

9) Mass density of Argon

Given:

T = -15.0°C = -15.0+273 = 258K

P= 10.0atm

V=1.00L

molar mass Argon = 39.95g/mol   (In tables)

According to gas law: PV=nRT

and n=moles = mass/molar mass

Combining both equations:

PV=RT(mass/molar mass) and

P(molar mass)/RT = mass/V

mass/V = (10.0atm)(39.95g/mol)/(0.0821Latm/molK)(258K) = 18.9g/L

18.9g/L (1L/1000ml) = 0.0189g/ml

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