Question

Question:A 4.50-kg block of ice at 0.00°C falls into the ocean and melts. The average temperature of the ocean is 3.50°C, including all the deep water. By how much does the change of this ice to water at 3.50o C alter the entropy of the world? Does the entropy increase or decrease? (Hint: Do you think that the ocean temperature will change appreciably as the ice melts?)

Short answer

The entropy of world increases by \(0.051\;{\rm{kJ}}/{\rm{K}}\).

Step-by-step solution

Identification of given data

The temperature for ice block is\({T_1} = 0\;^\circ {\rm{C}}\)

The average temperature ocean is\({T_2} = 3.50\;^\circ {\rm{C}}\)

The het input for engine is\({Q_1} = 1.5 \times {10^4}\;{\rm{J}}\)

The mass of ice block is \(m = 4.50\;{\rm{kg}}\)

Conceptual Explanation

The change in the entropy is due to latent heat to change of ice into water and rise of ice water temperature up to temperature of ocean.

Determination of change in entropy of world

The change in entropy of the surrounding is given as:

\(\Delta {s_1} = - \frac{{m\left( {L + c\left( {{T_2} - {T_1}} \right)} \right)}}{{{T_2}}}\)

Here,\(L\)is the latent heat of ice and its value is\(334\;{\rm{kJ}}/{\rm{kg}}\),\(c\)is the specific heat of water and its value is\(4.2\;{\rm{kJ}}/{\rm{kg}} \cdot {\rm{K}}\).

Substitute all the values in the above equation.

\(\begin{array}{l}\Delta {s_1} = - \frac{{\left( {4.50\;{\rm{kg}}} \right)\left( {334\;{\rm{kJ}}/{\rm{K}} + \left( {4.2\;{\rm{kJ}}/{\rm{kg}} \cdot {\rm{K}}} \right)\left( {\left( {3.50\;^\circ {\rm{C}} + 273} \right)\;{\rm{K}} - \left( {0\;^\circ {\rm{C}} + 273} \right)\;{\rm{K}}} \right)} \right)}}{{\left( {3.50\;^\circ {\rm{C}} + 273} \right)\;{\rm{K}}}}\\\Delta {s_1} = - 5.7\;{\rm{kJ}}/{\rm{K}}\end{array}\)

The change in entropy of the ocean is given as:

\(\Delta {s_2} = mc\ln \left( {\frac{{{T_2}}}{{{T_1}}}} \right)\)

Substitute all the values in the above equation.

\(\begin{array}{l}\Delta {s_2} = \left( {4.50\;{\rm{kg}}} \right)\left( {4.2\;{\rm{kJ}}/{\r\(\Delta s = \Delta {s_1} + \Delta {s_2} + \Delta {s_3}\)\(\begin{array}{l}\Delta {s_3} = \frac{{\left( {4.50\;{\rm{kg}}} \right)\left( {334\;{\rm{kJ}}/{\rm{K}}} \right)}}{{\left( {0\;^\circ {\rm{C}} + 273} \right)\;{\rm{K}}}}\\\Delta {s_3} = 5.51\;{\rm{kJ}}/{\rm{K}}\end{array}\)m{kg}} \cdot {\rm{K}}} \right)\ln \left( {\frac{{\left( {3.50\;^\circ {\rm{C}} + 273} \right)\;{\rm{K}}}}{{\left( {0\;^\circ {\rm{C}} + 273} \right)\;{\rm{K}}}}} \right)\\\Delta {s_2} = 0.241\;{\rm{kJ}}/{\rm{K}}\end{array}\)

The change in entropy of the ice is given as:

\(\Delta {s_3} = \frac{{mL}}{{{T_1}}}\)

Substitute all the values in the above equation.

The change in entropy of world is given as:

Substitute all the values in the above equation.

\(\begin{array}{l}\Delta s = - 5.7\;{\rm{kJ}}/{\rm{K}} + 0.241\;{\rm{kJ}}/{\rm{K}} + 5.51\;{\rm{kJ}}/{\rm{K}}\\\Delta s = 0.051\;{\rm{kJ}}/{\rm{K}}\end{array}\)

Therefore, the entropy of world increases by \(0.051\;{\rm{kJ}}/{\rm{K}}\).