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Chapter 15: Problem 48

An ice cube at \(0.0^{\circ} \mathrm{C}\) is slowly melting. What is the change in the ice cube's entropy for each \(1.00 \mathrm{g}\) of ice that melts?

Short Answer

Expert verified
Answer: The change in entropy for every 1.00 g of ice that melts from 0.0掳C is approximately 1.22 J/K.

Step by step solution

01

Determine the heat of fusion of ice

The heat of fusion is the amount of heat required to change a substance from the solid phase to the liquid phase without changing its temperature. For ice, the heat of fusion is equal to \(333.55 \mathrm{J/g}\).
02

Calculate the heat needed for melting 1 gram of ice

For every \(1.00 \mathrm{g}\) of ice that melts, we can determine the total heat needed for the process by multiplying the heat of fusion by the mass of ice. $$Q = m \times L = (1.00 \thinspace \mathrm{g}) \times (333.55 \thinspace \mathrm{J/g}) = 333.55 \thinspace \mathrm{J}$$
03

Calculate the change in entropy

To calculate the change in entropy, we will use the formula: $$\Delta S = \frac{Q}{T}$$ where \(\Delta S\) is the change in entropy, \(Q\) is the heat added to the system, and \(T\) is the temperature. Remember to use temperature in Kelvin. First, convert the temperature to Kelvin: $$T_{\text{K}} = T_{\text{C}} + 273.15 = 0.0^{\circ} \mathrm{C} + 273.15 \thinspace\mathrm{K} = 273.15 \thinspace\mathrm{K}$$ Now, calculate the change in entropy: $$\Delta S = \frac{333.55 \thinspace \mathrm{J}}{273.15 \thinspace \mathrm{K}} \approx 1.22 \thinspace \mathrm{J/K}$$ So, for every \(1.00 \mathrm{g}\) of ice that melts, the change in the ice cube's entropy is approximately \(1.22 \thinspace \mathrm{J/K}\).

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Most popular questions from this chapter

A \(0.500-\mathrm{kg}\) block of iron at \(60.0^{\circ} \mathrm{C}\) is placed in contact with a \(0.500-\mathrm{kg}\) block of iron at $20.0^{\circ} \mathrm{C} .\( (a) The blocks soon come to a common temperature of \)40.0^{\circ} \mathrm{C} .$ Estimate the entropy change of the universe when this occurs. [Hint: Assume that all the heat flow occurs at an average temperature for each block.] (b) Estimate the entropy change of the universe if, instead, the temperature of the hotter block increased to \(80.0^{\circ} \mathrm{C}\) while the temperature of the colder block decreased to \(0.0^{\circ} \mathrm{C}\) [Hint: The answer is negative, indicating that the process is impossible. \(]\)
(a) What is the entropy change of 1.00 mol of \(\mathrm{H}_{2} \mathrm{O}\) when it changes from ice to water at \(0.0^{\circ} \mathrm{C} ?\) (b) If the ice is in contact with an environment at a temperature of \(10.0^{\circ} \mathrm{C}\) what is the entropy change of the universe when the ice melts?
An inventor proposes a heat engine to propel a ship, using the temperature difference between the water at the surface and the water \(10 \mathrm{m}\) below the surface as the two reservoirs. If these temperatures are \(15.0^{\circ} \mathrm{C}\) and \(10.0^{\circ} \mathrm{C},\) respectively, what is the maximum possible efficiency of the engine?
A large block of copper initially at \(20.0^{\circ} \mathrm{C}\) is placed in a vat of hot water \(\left(80.0^{\circ} \mathrm{C}\right) .\) For the first $1.0 \mathrm{J}$ of heat that flows from the water into the block, find (a) the entropy change of the block, (b) the entropy change of the water, and (c) the entropy change of the universe. Note that the temperatures of the block and water are essentially unchanged by the flow of only \(1.0 \mathrm{J}\) of heat.
The United States generates about \(5.0 \times 10^{16} \mathrm{J}\) of electric energy a day. This energy is equivalent to work, since it can be converted into work with almost \(100 \%\) efficiency by an electric motor. (a) If this energy is generated by power plants with an average efficiency of \(0.30,\) how much heat is dumped into the environment each day? (b) How much water would be required to absorb this heat if the water temperature is not to increase more than \(2.0^{\circ} \mathrm{C} ?\)
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