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

answer : third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol

explanation :

let us write down the given equations

equation 1 :

equation 2 :

equation 3 :

let us use hess's law to find out which 2 equations can be added to get the third equation.

let us reverse equation 2 and multiply it by 2. the new equation that we get is,

equation 4 :

let us add equation 1 & 4

the above equation is similar to equation 3 but has different value for dhrxn.

therefore third trial had the contaminated sample and the correct value for dhrxn should be 1000 kj/mol