Substance A has the following properties.
mp at 1 atm: - 220.°C bp at 1 atm: 85°C
ΔHfus : 180. J/g ΔHvap : 500. J/g
Csolid : 1.0 J/g?°C C liquid : 2.5 J/g?°C
Cgas : 0.5 J/g?°C
At 1 atm, a 25-g sample of A is heated from 240.° C to 100.°C at a constant rate of 450. J/min. (a) How many minutes does it take to heat the sample to its melting point?



Answer :

It will take 1.11 min to heat the sample to its melting point.

Melting point = - 20°C

Boiling point = 85°C

∆H of fusion = 180 J/g

∆H of vap = 500 J/g

C(solid) = 1.0 J/g °C

C(liquid) = 2.5 J/g °C

C(gas) = 0.5 J/g °C

Mass of sample = 25 g

Initial temperature = - 40°C

Final temperature = 100°C

Rate of heating = 450 J/min

Specific heat capacity formula:- q = m ×C×∆T

Here, q = heat energy

        m = mass

        C = specific heat

      ∆T = temperature change

Melting point = - 20°C

C(solid) = 1.0 J/g °C

∆T = final temperature - initial temperature = -20 - (-40) = 20

Put these value in  Specific heat capacity formula

q = m ×C×∆T

q = 25×1.0×20

   =500J

The Rate of heating = 450 J/min

i.e. 450J = 1min

so, 500J = 1.11min

1.11 minutes does it take to heat the sample to its melting point.

The specific heat capacity is defined as the amount of heat absorbed in line with unit mass of the material whilst its temperature increases 1 °C.

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