Question: (I) 1.0 kg of water is heated from 0°C to 100°C. Estimate the change in entropy of the water.

Entropy is a function of the state of a system which is a measure of the order or disorder of a system.

When heat Q is added to a system by a reversible process, at a constant temperature T, then change in entropy of the system is

\(\Delta S = \frac{Q}{T}\).

The change in entropy of the steam can be calculated using this expression.

The mass of water is \(m = 1.0\;{\rm{kg}}\).

Temperature\({T_1}\)is\({T_1} = 0^\circ {\rm{C}} = \left( {0 + 273} \right)K = 273\;{\rm{K}}\).

Temperature\({T_2}\)is\({T_2} = 100^\circ {\rm{C}} = \left( {100 + 273} \right)K = 373\;{\rm{K}}\).

The specific heat of water is \(c = 4186\;{\rm{J/kg}} \cdot {\rm{K}}\).

Change in temperature of water is

\(\begin{aligned}{c}\Delta T &= {T_2} - {T_1}\\ &= \left( {373 - 273} \right)\;{\rm{K}}\\ &= 100\;{\rm{K}}{\rm{.}}\end{aligned}\)

The heat required to raise the temperature of unit mass of water by unit degree is termed the specific heat of water.

The net heat required to raise the temperature is

\(\begin{aligned}{c}\frac{Q}{{m\Delta T}} &= c\\Q &= mc\Delta T.\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}{c}Q &= \left( {1.0\;{\rm{kg}}} \right) \times \left( {4186\;{\rm{J/kg}} \cdot {\rm{K}}} \right) \times \left( {100\;{\rm{K}}} \right)\\ &= 4,18,600\;{\rm{J}}\end{aligned}\)

This is the amount of heat given to the water to change its temperature.

Average temperature of water is

\(\begin{aligned}{c}T &= \frac{{{T_1} + {T_2}}}{2}\\ &= \frac{{\left( {273 + 373} \right)\;{\rm{K}}}}{2}\\ &= 323\;{\rm{K}}{\rm{.}}\end{aligned}\)

The change in entropy of the water is

\(\Delta S = \frac{Q}{T}\).

Substitute the values in the above expression.

\(\begin{aligned}{c}\Delta S &= \frac{{418,600\;{\rm{J}}}}{{323\;{\rm{K}}}}\\ &= 1295.98\;{\rm{J/K}}\\ &= 1.3 \times 1{{\rm{0}}^3}\;{\rm{J/K}}\end{aligned}\)

Thus, change in entropy of the water is \(1.3 \times 1{{\rm{0}}^3}\;{\rm{J/K}}\).