A certain objects\' temperature increases by 1.0 degree C for every 1560 J that

Question

A certain objects' temperature increases by 1.0 degree C for every 1560 J that it gains. A 0.1964 g sample of quinone (molar mass = 108.1 g mole) was burnt in it, and the surrounding object's temperature increased from 20.3 degree C to 23.5 degree C. Find the molar enthalpy of combustion of quinone. If the pressure is very low, ice can sublimate at -70 degree C. instead of melting and then evaporating. To calculate the enthalpy of sublimation (Delta H_subl) at such temperature, we can take advantage of the properties of state functions. In the T vs. H diagram below, draw a plausible path to go from H_2O (s) at -70degree C to H_2O (g) at the same temperature. H_2O (s) H_2O (g) Write the algebraic expression to calculate Delta H^-70_subl for the path you drew in the diagram.

Explanation / Answer

21)

specific heat = 1560 J/ oC

Q = Cp dT

     = 1560 x (23.5 - 20.3 )

     = 4992 J

moles of quinone = 0.1964 / 108.1 = 1.817 x 10^-3

delta H = - Q / n

             = - 4992 x 10^-3 / 1.817 x 10^-3

delta H = 2747.6 kJ/mol

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