A 41.81 g sample of a substance is initially at 20.2 °C. After absorbing 1375 J of heat, the temperature of the substance is 184.6 °C. What is the specific heat (c) of the substance?

c=0.2 J/g°C

Explanation:

Given that,

Mass, m = 41.81 g

Initial temperature,

Final temperature,

The heat absorbed by the substance is 1375 J.

We need to find the specific heat (c) of the substance. It can be given by the below formula:

Hence, the specific heat (c) of the substance is 0.2 J/g°C.

1 a

2 d

3 a

4 c

explanation:

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