A 488.3 gram sample of an unknown substance (MM = 92.41 g/mol) is heated from -23.1 °C to 51.8 °C. (heat capacity of solid = 2.96 J/g･°C; heat capacity of liquid = 1.75 J/g･°C; ∆Hfus = 8.04 kJ/mol; Tfinal = 17.6 °C) a) How much energy (in kJ) is absorbed/released to heat the solid?
b)How much energy (in kJ) is absorbed/released to melt the solid?
c)How much energy (in kJ) is absorbed/released to heat the liquid?
d) What is the total amount of energy that must be absorbed/released for the entire process?

a) Q₁ = 58.82 KJ

b) Q₂ = 42.48 KJ

c) Q₃ = 29.22 KJ

d) Q = 130.52 KJ  

Explanation:

a)

In order to find the energy absorbed to heat the solid, we will use:

where,

Q₁ = Heat absorbed for heating solid = ?

m = mass of solid = 488.3 g = 0.4883 kg

C₁ = Specific Heat Capacity of Solid = 2.96 J/g °C

ΔT₁ =  Change in temperature of Solid = Melting Temperature - Initial Temp.

ΔT₁ = 17.6°C - (-23.1°C) = 40.7°C

Therefore,

Q₁ = 58.82 KJ

b)

In order to find the absorbed to melt the solid at 17.6°C, we will use:

where,

Q₂ = Heat absorbed for melting solid = ?

H_fus = Heat of Fusion = 8.04 KJ/mol

n = no. of moles =

Therefore,

Q₂ = 42.48 KJ

c)

In order to find the energy absorbed to heat the liquid, we will use:

where,

Q₃ = Heat absorbed for heating Liquid = ?

m = mass of solid = 488.3 g = 0.4883 kg

C₃ = Specific Heat Capacity of Liquid = 1.75 J/g °C

ΔT₃ =  Change in temperature of Liquid = Final Temp. - Melting Temp.

ΔT₃ = 51.8°C - 17.6°C = 34.2°C

Therefore,

Q₃ = 29.22 KJ

d)

Total amount of energy absorbed during entire process is:

Q = 130.52 KJ

yes, water is wet, just as it makes other things wet, it is wet itself, kinda like ice, when ice is in water, the ice makes water cold, and ice is cold, therefore with that example, water is wet

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