Calculate the specific heat of a substance if a 35g sample absorbs 48 j as the temperature is raised from 293 k to 313 k

c = 0.07 j/g.k

Explanation:

Given data:

Mass of sample = 35 g

Heat absorbed = 48 j

Initial temperature = 293 K

Final temperature = 313 K

Specific heat of substance = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

ΔT = 313 k - 293 K

ΔT = 20 k

Now we will put the values in formula.

48 j = 35 g × c× 20 k

48 j = 700 g.k ×c

c = 48 j/700 g.k

c = 0.07 j/g.k

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.

d.) the average kinetic energy of the gas particles is greater in container b because its particles move faster.