Example: For a coal-fire electric power plant, T hot(the boiler temperature)would be 825 K, and Tcold(the cooling tower)would be about 300 K. This leads to Efficiency=(1-300/825)X100%=(1-0.36)×100%=64% In this case, 36% of the heat energy from the energy of the fuel must be wasted by rejecting it through the cooling tower to the surrounding atmosphere To make the efficiency as high as possible it would be desirable to increase T hot and decrease The limit on Thot is imposed by the materials from which the boilers can be constructed and the limit on Tcold is imposed by the availability in nature of large sinks at sufficiently low temperatureExample: For a coal-fire electric power plant, Thot (the boiler temperature) would be 825 K, and Tcold (the cooling tower) would be about 300 K. This leads to Efficiency = (1 – 300 / 825) x 100% = (1 – 0.36) x 100% = 64% • In this case, 36% of the heat energy from the energy of the fuel must be wasted by rejecting it through the cooling tower to the surrounding atmosphere. • To make the efficiency as high as possible, it would be desirable to increase Thot and decrease Tcold. • The limit on Thot is imposed by the materials from which the boilers can be constructed and the limit on Tcold is imposed by the availability in nature of large sinks at sufficiently low temperature