  We derive an analytical expression for a critical temperature of spontaneous
symmetry breaking in a repulsive hard-core interacting Bose system. We show
that the critical temperature of spontaneous symmetry breaking in a hard-core
interacting Bose system is determined by the three physical parameters: the
density of Bose liquid at absolute zero (rho_0), the mass (m) and the hard
sphere diameter (sigma) of a boson. The formula that we have derived is Tc =
rho_0 pi hbar^2 sigma / m^2 kB. We report Tc of liquid helium-4 is 2.194 K,
which is significantly close to the lambda temperature of 2.1768 K. The
deviation between the predicted and experimental values of the lambda
temperature is less than 1%.
