One gram of steam at hundred degree Celsius melts how much gram of ice at zero degree Celsius?
A. One gram
B. 2 gram
C. 4 gram
D. 8 gram

Hint: Calculate the heat required to change the state of steam into water at the same temperature, then calculate the heat energy loss as the temperature of steam decreases. According to the principle of calorimetry, the total heat given is equal to total heat taken.

Formula used:
\[\Delta Q = mc\Delta T\]
Here, m is the mass and c is the specific heat.

Complete step by step answer:
The heat required to convert steam at \[100^\circ C\] to water at \[100^\circ \] is can be calculated as,
\[\Delta Q = {m_{steam}}{L_{steam}}\]

The heat lost by the water at \[100^\circ \] to cool it to \[0^\circ \] is,
\[\Delta Q = {m_w}{C_w}\Delta T\]

Here, \[{m_w}\] is the mass of water, \[{C_w}\] is the specific heat of water and \[\Delta T\] is the change in temperature.

Therefore, the total heat given to ice is,
\[{m_{steam}}{L_{steam}} + {m_w}{C_W}\Delta T\]

The heat required to melt ice of mass \[{m_{ice}}\] is,
\[\Delta Q = {m_{ice}}{L_{ice}}\]

Here, \[{L_{ice}}\] is the latent heat of ice.

According to the principle of calorimetry, the heat given is equal to heat taken. Therefore,
\[{m_{steam}}{L_{steam}} + {m_w}{C_W}\Delta T = {m_{ice}}{L_{ice}}\]
\[ \Rightarrow {m_{ice}} = \dfrac{{{m_{steam}}{L_{steam}} + {m_w}{C_W}\Delta T}}{{{L_{ice}}}}\]

Substitute 1 gram for \[{m_{steam}}\], \[540\,cal/g\] for \[{L_{steam}}\], 1 gram for \[{m_w}\], 1 cal/g for \[{C_W}\], and 80 cal/g for \[{L_{ice}}\] in the above equation.
\[{m_{ice}} = \dfrac{{\left( 1 \right)\left( {540} \right) + \left( 1 \right)\left( 1 \right)\left( {100^\circ - 0^\circ } \right)}}{{80}}\]
\[ \Rightarrow {m_{ice}} = 8\,g\]

Therefore, the mass of ice that can be melted by the steam is 8 gram.

So, the correct answer is option (D).

Note:The heat required to convert the steam into water is known as heat of sublimation and it is given by the product of mass of the substance and latent heat of sublimation of the substance \[mL\]. The water at \[100^\circ C\] consists of heat given by the expression \[mc\Delta T\]. Therefore, the total heat given by the steam is the sum \[mL + mC\Delta T\]. The temperature in the above formula should be in the Kelvin scale but since it is the difference in the temperature, it will be the same for both Celsius and Kelvin scale.