91影视

When different parts of an isolated system are at different temperatures, heat energy flows from the part which is at a higher temperature to the part which is at a lower temperature until thermal equilibrium is reached.

Thus, according to the law of conservation of energy, 鈥淔or an isolated system, the heat lost by one part of the system must be equal to the heat gained by the other part.鈥�

Heat lost = Heat gained

Mass of ice cube is \({m_1} = 28\;{\rm{g}} = 28 \times {\rm{1}}{{\rm{0}}{ - 3}}\;{\rm{kg}}\).

The temperature of ice is \({T_1} = 0\circ {\rm{C}} = \left( {0 + 273} \right)\;{\rm{K}} = 2{\rm{73}}\;{\rm{K}}\).

The temperature of nitrogen is \({T_2} = 77\;{\rm{K}}\).

Specific heat of ice is \({c_{{\rm{ice}}}} = 2100\;{\rm{J/kg}} \cdot {\rm{K}}\).

Latent heat of vaporization of nitrogen is \({L_{\rm{V}}} = 200\;{\rm{kJ/kg}} = 200 \times {103}\;{\rm{J/kg}}\).

When the ice cube is dropped into the liquid nitrogen, the amount of heat lost by ice to change its temperature from \({T_1}\) to \({T_2}\) is given by

\({Q_{{\rm{Lost}}}} = {m_1}{c_{{\rm{ice}}}}\left( {{T_1} - {T_2}} \right)\).

The amount of heat gained by liquid nitrogen to evaporate is given by

\({Q_{{\rm{gained}}}} = m{L_{\rm{V}}}\).

Here, m is the mass of the liquid nitrogen.

Apply the law of conservation of heat energy.

\(\begin{array}{c}{Q_{{\rm{gained}}}} = {Q_{{\rm{Lost}}}}\\m{L_{\rm{V}}} = {m_1}{c_{{\rm{ice}}}}\left( {{T_1} - {T_2}} \right)\\m = \frac{{{m_1}{c_{{\rm{ice}}}}\left( {{T_1} - {T_2}} \right)}}{{{L_{\rm{V}}}}}\end{array}\)

Substitute the values into the above expression.

\(\begin{array}{c}m = \frac{{\left( {28 \times {{10}{ - 3}}\;{\rm{kg}}} \right)\left[ {\left( {2100\;{\rm{J/kg}} \cdot {\rm{K}}} \right)\left( {273 - 77} \right){\rm{K}}} \right]}}{{200 \times {{10}3}\;{\rm{J/kg}}}}\\ = 57.624 \times {10{ - 3}}\;{\rm{kg}}\\ = 57.624\;{\rm{g}}\end{array}\)

Thus, the amount of nitrogen that evaporates is \(57.624\;{\rm{g}}\).