The melting point of a certain substance is 70°C, its normal boiling point is 450°C, its enthalpy of fusion is 30 cal/g, enthalpy of vaporization is 45 cal/g, and its specific heat is 0.215 cal/g-K. T

Question

The melting point of a certain substance is 70°C, its normal boiling point is 450°C, its enthalpy of fusion is 30 cal/g, enthalpy of vaporization is 45 cal/g, and its specific heat is 0.215 cal/g-K. The heat required (in cal) to convert 10 g of the substance from the solid state at 70°C to vapour at 450°C is

Answer:

1567.00

Solution:

Melt: 10×30 = 300 cal. Heat liquid 70 to 450°C: 10×0.215×380 = 817 cal. Vaporize: 10×45 = 450 cal. Total = 300+817+450 = 1567 cal. Ans = 1567.

TestHub · Accepted Answer

Melt: 10×30 = 300 cal. Heat liquid 70 to 450°C: 10×0.215×380 = 817 cal. Vaporize: 10×45 = 450 cal. Total = 300+817+450 = 1567 cal. Ans = 1567.

Chemistry - Thermodynamics - I Question with Solution | TestHub

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