How much heat is added if .7892g of water is increased in temperature by .039 degrees c?

=0.1293 J

Explanation:

The heat increase is the change in energy for the water ΔH.

ΔH=MC∅ where M is mass, C is the specific heat capacity for water and ∅ is the change in temperature.

M=0.7892g

C(for water) =4.2 J/g°C

∅=0.039 °C

ΔH=0.7892g×4.2 J/g°C×0.039°C

=0.1293 J

The amount of heat energy added to cause the 0.039°C temperature increase is 0.1293 Joules.

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