Suppose you need to calibrate a calorimeter. You burn 0.849 g of a fuel with an enthalpy of combustion of − 48.7 kJ/g . If the temperature of water in the calorimeter increases by 11.4 ∘ C during the reaction, what is the heat capacity of the calorimeter in kJ/C°?

the transfer of heat by electromagnetic radiation is called thermal radiation . it is usually carried out by electromagnetic waves in an energy range just below the visible spectrum, which is the infrared range.

so the answer would be radiation.

: )

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole