How many joules of heat are absorbed to raise the temperature of 125 grams of water at 1 atm from 35°C to its boiling point, 100.°C.

33962.5 J

Explanation:

From the question given above, the following data were obtained:

Mass (M) = 125 g

Initial temperature (T₁) = 35 °C

Initial temperature (T₂) = 100 °C

Specific heat capacity (C) = 4.18 J/gºC

Heat (Q) absorbed =?

Next, we shall determine the change in temperature. This can be obtained as follow:

Initial temperature (T₁) = 35 °C

Initial temperature (T₂) = 100 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 100 – 35

ΔT = 65 °C

Finally, we shall determine the heat absorbed. This can be obtained as follow:

Mass (M) = 125 g

Change in temperature (ΔT) = 65 °C

Specific heat capacity (C) = 4.18 J/gºC

Heat (Q) absorbed =?

Q = MCΔT

Q = 125 × 4.18 × 65

Q = 33962.5 J

Therefore, the heat absorbed is 33962.5 J

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