A coffee cup calorimeter contains 153.21 g of water at 22.50 °C. A 65.454 g piece of iron is heated to 101.62 °C. The piece of iron is added to the coffee cup caloriemter and the contents reach thermal equilibrium at 25.68 °C. The specific heat capacity of iron is 0.449 Jg·K and the specific heat capacity of water is 4.184 Jg·K. How much heat, , is lost by the piece of iron?

Q = 2232J

Explanation:

In the system, the energy that is lost by the iron is:

Q = S*ΔT*m

Where Q is heat, S is specific heat of iron ΔT is change in temperature and m is mass of the piece of iron.

S = 0.449J/gK

ΔT = 101.62°C - 25.68°C = 75.94°C

m = 65.454g

Q = 0.449J/gK*75.94°C*65.454g

colorless gas, exhibiting a blue glow when placed in an electric field

b) solid

its a type of rock

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