To determine the quality of thermal insulation of a Dewar vessel, it is filled with ice at`0^(@)C` 42 g of ice have melted in 24 h. Usually liquid nitrogen at 78 K is kept in this flask. Assuming the quantity of heat entering the flask to be proportional to the difference in the internal and the external temperatures of the vessel, find the amount of liquid nitrogen that is going to evaporate in 24 h. The ambient temperature is `20^(@)C` the heat of vapourization of liquid nitrogen at normal pressure is `1.8xx10^(5)J//K`.

Explanation

The heat flowing to the Dewar vessel is `Q=alpha(T_("air")-T)`, . where `alpha` is a certain coefficient, and T is the temperature inside the flask. For ice and liquid nitrogen we obtain the ratio
`(Q_(1))/(Q_(2))=(T_("air")T_(1))/(T_("air")T_(2))`
But for nitrogen `Q_(2)=m_(2)L` where L is its heat of evaporation, the respective value for ice being `Q_(1)=m_(1) lambda`. Hence
`(m_(1)lambda)/(m_(2)L)=(T_("air")-T_(1))/(T_("air")-T_(2))`, from which `m_(2)=(m_(1) lambda(T_("air")-T_(2)))/(L(T_("air")-T_(1)))`