Answer:
1716.475 J/kg.°C.
Explanation:
Specific Heat capacity is given as
Q = cm(t₂-t₁) ................ Equation 1
Where Q = quantity of heat, c = specif heat of benzene, m = mass of benzene, t₁ = initial temperature, t₂ = final temperature.
Make c the subject of the equation,
c = Q/m(t₂-t₁) ................. Equation 2
Given: Q = 11.2 kJ = 11200 J m = 145 g = 0.145 kg, t₂ = 68.0°C, t₁ = 23.0°C
Substitute into equation 2.
c = 11200/[0.145(68-23)]
c = 11200/6.525
c = 1716.475 J/kg.°C.
Hence the specific heat of benzene = 1716.475 J/kg.°C.
