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A 34.1-mL sample of benzene at 20.8°C was cooled to its melting point, 5.5 °C, and then frozen at 5.5 °C. Calculate the quantity of energy transferred to the surroundings in this process. The density of benzene is 0.876 g/mL. The specific heat capacity of the liquid is 1.74 J g-1 K-1. Its fusion enthalpy is 127 J/g. J

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mnhlabatsi7

Answer:

4541.8 J

Explanation:

First we find the mass of benzene available

mass = density x volume

         = 0.867 x 34.1

         = 29.5647 g

Then we find the amount of heat transferred by two processes:

heat tranferred = heat lost during temp drop + heat lost during freezing

                         = mcΔT + mL

                         = 29.5647 x 1.74 x (20.8 - 5.5) + 29.5647 x 127

                         = 4541.7883434 J

                         = 4541.8 J

                         

greenkelvin5

Answer:

4541.8 J or 4.542 KJ

Explanation:

From our knowledge of density of a substance,  density=mass/volume

∴ mass=density×volume ⇒0.876×34.1=29.565 g

The quantity of heat transferred to the surrounding ∈ΔH=mcΔT+ML

=29.565×1.74(20.8-5.5)+(29.565× 127) = 4541.8 J or 4.542 KJ

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