Question

Mercury melts at 234K and boils at 298K at 2x10^-3 torr of pressure. Heat is added to a sample containing 3.00 kg of mercury (Hg) at 238K. The rate of heating is 2000 watts. The pressure is constant and you can assume table values are not affected by reduced pressure or temperature. At what time will the mercury start to boil? How much time for the mercury to completely boil? How long until the mercury gas heats up to 498K?

Answer 1

Given Power, P = Q/t(s) = 2000 W

=> Q = 2000xt(s)

Specific heat capacity of Hg(liquid), S = 1250 JKg^-1DegC^-1

(a) Mass of mercury, m = 3.00 Kg

Initial temperature, Ti = 238K

Mercury will start to boil at 298K and 2x10-3 torr of pressure, hence final temperature, Tf = 298K

Heat requird, Q = mxSxdT = 3.00Kg x (1250 JKg^-1DegC^-1)x(298K - 238K) = 225000 J

Also Q = 2000xt(s) = 225000 J

=> t(s) = 225000J / 2000W = 112.5 s

Note: Use the value of S provided in your book to get exact answer.

Hence after 112.5 s the mercury will start to boil.

(b): Latent heat of Vaporization of mercury = 295000 JKg^-1

Hence latent heat of vaporization for 3.00 Kg of mercury, Hv = 3Kg x ( 295000 JKg^-1) = 885000J

To boil the mercury completely, heat equivalent to Hv must have to be given. Hence

total heat requird, Q = mxSxdT + Hv = 3.00Kg x (1250 JKg^-1DegC^-1)x(298K - 238K) + 885000J = 1110000 J

Also Q = 2000xt(s) = 1110000J

=> t(s) = 1110000J / 2000W = 555 s (answer)

Hence after 555 s the mercury will completely boil.

(c):

Specific heat capacity of Hg(gas), Sg = 1040JKg^-1DegC^-1 Ti = 238K, Tf = 498K

total heat requird, Q = mxSxdT + Hv + mSgxdT

= 3.00Kg x (1250 JKg^-1DegC^-1)x(298K - 238K) + 885000J + 3.00Kg x (1040 JKg^-1DegC^-1)x(498K - 298K)

= 1734000J

Also Q = 2000xt(s) = 1734000J

=> t(s) = 1734000J / 2000W = 867 s (answer)

Hence after 867 s the mercury will heat upto 498K