Draw heating curves and then calculate q for the following processes (given that

Question

Draw heating curves and then calculate q for the following processes (given that substance X melts at 27.2 degree C, boils at 99.7 degree C and has the following specific heat capacities; C_solid = 2.1 J/g/degree C, c_liquid = 2.7 J/g/ degree C, C_vapor = 1.4 J/g/ degree C, Delta H_fus = 923 J/g, Delta H_vap = 2016 J/g,); (a) 8.66 g of substance X changing from solid at 27.2 degree C to liquid at 27.2 degree C. (b) 189 mg of substance X changing from 86 degree C to 14.4 degree C (c) 2 kg of substance X changing from -24.8 degree C to 151 degree C

Explanation / Answer

a)

this is melting, so

Qtotal = m*Hmelting

Qtotal = 8.66 g * 923 J/g = 7993.18 J

b)

Clearly, it goes from liquid to solid, sicne T= 27.2°C is in between

so

Qtotal = Qliquid from 86° to 27.2° + Qmelting + Qsolid from 27.2°C to 14.4°C

calculate

Qliquid from 86° to 27.2° = m*Cl*(Tf-Ti) = 0.189 g* 2.7 J/gC* (27.2-86)C= - 30.00564

Qmelting = 0.189g * -923 J/g = -174.447 J

Qsolid from 27.2°C to 14.4°C = m*C*(Tf-Ti) = 0.189*2.1*(14.4-27.2) = - 5.08032 J

total Q = -(30.00564+174.447 +5.08032 ) = -209.53296 J

c=

m = 2kg = 2000 g

once again, it goes form solid to liquid, then liquid to solid at T = 99.7°C

Qtotal = Qsolid from -24.8° to 27.2° + Qfreezing + Qliquid from 27.2°C to 99.7°C + Qvapor + Qgas from 99.7 to 151

Qtotal = (2000)(2.1)(27.2--24.8) + 2000*923 + 2000*2.7*(99.7-27.2) + 2000*2016 + 2000*1.4*(151-99.7)=

Qtotal = 6631540 j

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