91Ó°ÊÓ

When the phase of a material changes from solid to liquid or from liquid to gas, the material absorbs some amount of heat energy. The amount of heat required to change the phase of a material of unit mass at constant temperature is termed latent heat.

The amount of heat required to change the phase of 1.0 kg of a substance from solid to liquid is termed latent heat of fusion \(\left( {{L_{\rm{F}}}} \right)\). Its value for ice is \({L_{\rm{F}}} = 3.33 \times {105}\;{\rm{J/kg}}\).

Mass of snow is \(m = 1.0\;{\rm{kg}}\).

The temperature of snow is \({T_1} = - 15\circ {\rm{C}}\).

The melting point of ice is \(T = 0\circ {\rm{C}}\).

The temperature of the water is \({T_2} = 2\circ \;{\rm{C}}\).

The temperature of your body is \({T_3} = 37\circ \;{\rm{C}}\).

Specific heat of ice is \(c = 2100\;{\rm{J/kg}} \cdot {\rm{\circ C}}\).

Specific heat of water is \(c' = 4186\;{\rm{J/kg}} \cdot {\rm{\circ C}}\).

Latent heat of fusion is \({L_{\rm{F}}} = 3.33 \times {105}\;{\rm{J/kg}}\).

When you eat 1.0 kg of snow, the temperature of ice increases from \(\left( {{T_1}} \right)\) to its melting point T by absorbing heat \(\left( { = mc\left( {T - {T_1}} \right)} \right)\) from your body; it then converts into water at temperature T by gaining heat \(\left( { = m{L_{\rm{V}}}} \right)\), and ultimately, the temperature of this water rises to reach the body temperature \(\left( {{T_3}} \right)\) by gaining the heat \(\left( { = mc'\left( {{T_3} - T} \right)} \right)\) from the body.

The total heat energy absorbed from the body is

\(Q = mc\left( {T - {T_1}} \right) + m{L_{\rm{F}}} + mc'\left( {{T_3} - T} \right)\).

Substitute the values into the above expression.

\(\begin{array}{c}Q = \left[ {\left( {1.0\;{\rm{kg}}} \right)\left( {2100\;{\rm{J/kg}} \cdot {\rm{\circ C}}} \right)\left( {0 - \left( { - 15} \right)} \right)\circ {\rm{C}}} \right] + \left[ {\left( {1.0\;{\rm{kg}}} \right)\left( {3.33 \times {{10}5}\;{\rm{J/K}}} \right)} \right]\\ + \left[ {\left( {1.0\;{\rm{kg}}} \right)\left( {4186\;{\rm{J/kg}} \cdot {\rm{\circ C}}} \right)\left( {37 - 0} \right)\circ {\rm{C}}} \right]\\ = 519,382\;{\rm{J}}\\ \approx 5.2 \times {105}\;{\rm{J}}\end{array}\)

Thus, the energy absorbed from the body when you eat 1.0 kg snow is \(5.2 \times {105}\;{\rm{J}}\).

When you drink 1.0 kg of melted ice or the water at temperature \({T_2}\), the temperature of this water rises from \({T_2}\) to the body temperature \({T_3}\) by absorbing heat \(\left( { = mc'\left( {{T_3} - {T_2}} \right)} \right)\) from the body.

The heat energy absorbed from the body is

\(Q = mc'\left( {{T_3} - {T_2}} \right)\).

Substitute the values into the above expression.

\(\begin{array}{c}Q = \left( {1.0\;{\rm{kg}}} \right)\left( {4186\;{\rm{J/kg}} \cdot {\rm{\circ C}}} \right)\left( {37 - 2} \right)\circ {\rm{C}}\\ = 146,510\;{\rm{J}}\\ \approx 1.5 \times {105}\;{\rm{J}}\end{array}\)

Thus, the energy absorbed from the body when you drink 1.0 kg of water is \(1.5 \times {105}\;{\rm{J}}\), which is significantly less than the heat energy absorbed on eating 1.0 kg of ice.