A 6.00 gram sample of solid gold was heated from 274 K to 325 K. How much energy was involved? Cs of gold is 0.129 J/g oC.

39.5J

Explanation:

Given parameters:

Mass of solid gold  = 6g

Initial temperature  = 274K

Final temperature  = 325K

Specific heat capacity of gold  = 0.129J/g°C

Unknown:

Amount of energy used  = ?

Solution:

The amount of energy absorbed in changing in giving a temperature rise to a body is given as a product of its mass, specific heat capacity and change in temperature.

Mathematically;

            H  = m x C x Ф

where m is the mass of the body

            c is the specific heat capacity

            Ф is the change in temperature

Now the unknown is H;

  We need to convert  the given temperatures into °C from kelvin so we can work with it;

     Initial temperature = 274K, in °C;       274 - 273 = 1°C

     Final temperature  = 325K, in °C;       325 - 273 = 52°C

Now input the parameters and solve;

  H   =  6 x 0.129 x (52 - 1) = 39.5J

For the reaction 2 K + F2 --> 2 KF,
consider K atomic wt. = 39
23.5 g of K = 0.603 moles, hence following the molar ratio of the balanced equation, 0.603 moles of potassium will use 0.3015 moles of F2. (number of moles, n = 0.3015)

Now, following the ideal gas equation, PV = nRT
P = 0.98 atm
V = unknown
n = 0.3015 moles
R = 82.057 cm^3 atm K^-1mole^-1 (unit of R chosen to match the units of other parameters; see the reference below)
T = 298 K
Solving for V,
V = (nRT)/P = (0.3015 mol * 82.057 cm^3 atm K^-1 mol^-1 * 298 K)/(0.98 atm)
solve it to get 7517.6 cm^3 as the volume of F2 = 7.5176 liters of F2 gas is needed.