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_) 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 - 70 degree C to H_2O (g) at the same temperature. Write the algebraic expression to calculate Delta H^-70 _subt for the path you drew in the diagram.

Explanation / Answer

Temperature change = 23.5-20.3 = 3.2 oC

amount of heat gained = 3.2oC * 1560J = 4992 J

Assume that heat liberated on burning = heat gained by the object

Moles of quinone = 0.1964/108.1g/mol = 1.82*10^-3

molar enthalpy = 4992J/ 1.82*10^-3 = 2747.63 kJ/mol

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