Question

A certain substance has a mass of 50/gmol. When 300 J of heat is added to 25 g of sample of this material, its tempertuare rises from 25 to `45^(@)`C. Calcualte (i) thermal capacity.. (ii) Specific heat and (iii) molar heat capacity of the sample.

Answer 1

Given, Mass m = 25g,
Temperature `DeltaT` = 45-25 = `20^(@)`C,
Heat `DeltaQ` = 300J ltbr Molar mass M = 50g/mol
(i) Thermal capacity, S`=(DeltaQ)/(DeltaT) = (300/20)` = 15`J//^(@)`C
(ii) Specific heat c= `(DeltaQ)/(DeltaT xx m) = 15 xx 1/25 = 0.6 `J//g^(@)C`
(iii) Number of moles n= (Mass(m))/(Molecualr mass (M)) = `25/50` = `1/2`
`therefore` `DeltaQ = nCDeltaT`
`therefore` Molar heat capacity, C = `(DeltaQ)/(nDeltaT)=(300)/((1/2)(20))= 30J//mole^(@)`C