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Chapter 19: Problem 64

A \(500-\mathrm{g}\) copper block at \(80^{\circ} \mathrm{C}\) is dropped into \(1.0 \mathrm{~kg}\) of water at \(10^{\circ} \mathrm{C}\). Find (a) the final temperature and (b) the entropy change of the system.

Short Answer

Expert verified
The final temperature of the system is \(14.9 \, 掳C\) and the entropy change of the system is \(0.062 \, J/K\).

Step by step solution

01

Calculate Heat Lost or Gained

Because the system is closed, the heat gained by water \(q_{water}\) is equal to the heat lost by copper \(q_{copper}\). From the specific heat formulas, we can then write them as the following equations:1. \(q_{water} = m_{water} * c_{water} * (T_{f} - T_{water})\)2. \(q_{copper} = m_{copper} * c_{copper} * (T_{copper} - T_{f})\)where \(m\) is mass, \(c\) is specific heat, \(T_{f}\) is the final temperature, and \(T\) is the initial temperature. The specific heat of water is \(4.18 \, J/g掳C\) and of copper is \(0.385 \, J/g掳C\).
02

Calculating the Final Temperature

By setting the heat lost by copper equal to the heat gained by water, we obtain the final temperature of the system \((T_{f})\): \(m_{water} * c_{water} * (T_{f} - T_{water}) = m_{copper} * c_{copper} * (T_{copper} - T_{f})\)Solving the equation for \(T_{f}\), substituting the given values and calculating, we find \(T_{f} = 14.9 \, 掳C\).
03

Calculate the Entropy Change

The entropy change \(\Delta S\) is calculated using the formula \(\Delta S = \frac{Q}{T}\), where \(Q\) is the heat transferred and \(T\) is the absolute temperature. This should be calculated separately for both the copper block and the water, then added together to get the total entropy change:1. \(\Delta S_{water} = \frac{q_{water}}{T_{water}}\)2. \(\Delta S_{copper} = \frac{q_{copper}}{T_{copper}}\)Therefore, the total entropy change \(\Delta S_{total} = \Delta S_{water} + \Delta S_{copper}\). Converting temperatures to Kelvins, substituting the previously calculated values and calculating, we get \(\Delta S_{total} = 0.062 \, J/K\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heat Transfer
Heat transfer is the process of energy moving from one body or substance to another. In thermodynamics, heat transfer can occur through conduction, convection, or radiation. In our exercise, conduction is the primary form of heat transfer. The copper block, initially at a higher temperature, transfers its heat energy to the cooler water until thermal equilibrium is reached. This means both substances reach the same final temperature. Key points to remember about heat transfer:
  • Heat moves from warmer to cooler objects.
  • In a closed system, the total energy (heat) remains constant.
  • The goal is to achieve thermal equilibrium.
Understanding how heat moves in this scenario helps us calculate significant quantities like final temperature and entropy.
Specific Heat Capacity
Specific heat capacity is a property that describes how much heat energy a substance needs to change its temperature by a certain amount. It's vital to know that different substances have different specific heat capacities. For this problem:
  • The specific heat capacity of water is about 4.18 J/g掳C.
  • The specific heat capacity of copper is 0.385 J/g掳C.
This means water needs more energy to change its temperature compared to copper. Thus, when the copper block and water are placed together, this property helps us determine how much heat each will gain or lose for them to reach the same final temperature, which brings us to the next part of our exercise.
Entropy Change
Entropy is a measure of disorder or randomness within a system. In thermodynamics, it's crucial for understanding how energy is distributed. When heat is transferred in a system, like with our copper and water, entropy will change.To calculate entropy:
  • Use the formula: \(\Delta S = \frac{Q}{T}\)
  • Calculate separately for each component (e.g., copper and water).
  • Combine these to find the total entropy change.
In this exercise, we calculated an entropy change of 0.062 J/K after converting temperatures to Kelvin and using our heat transfer values. Greater entropy changes can indicate less energy available for work.
Final Temperature Calculation
The final temperature is when both substances in the system reach thermal equilibrium. Solving for this requires knowing the initial temperatures, masses, and specific heat capacities, combined with the principle of heat conservation (heat lost equals heat gained).Steps to calculate it:
  • Set the heat gained by water equal to the heat lost by the copper block.
  • Formulate the equation with specific heat formulas.
  • Solve for the final temperature \(T_f\).
Using the provided values, the calculation gives a final temperature of 14.9掳C. This result stems from balancing the higher initial energy of the copper with the larger heat capacity of the water.

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Most popular questions from this chapter

A \(6.0-\mathrm{mol}\) sample of ideal diatomic gas is at \(1.00 \mathrm{~atm}\) pressure and \(350 \mathrm{~K}\). Find the entropy change as the gas is heated reversibly to \(510 \mathrm{~K}\) (a) at constant volume, (b) at constant pressure, and (c) adiabatically.

A Carnot engine extracts \(745 \mathrm{~J}\) from a \(592-\mathrm{K}\) reservoir during each cycle and rejects \(458 \mathrm{~J}\) to a cooler reservoir. It operates at \(18.6\) cycles per second. Find (a) the work done during each cycle, (b) its efficiency, (c) the temperature of the cool reservoir, and (d) its mechanical power output.

A Carnot engine's mechanical power output is \(48.4 \mathrm{~kW}\), and it rejects heat to the ambient environment at the rate of \(41.7 \mathrm{~kW}\). If the engine's hot reservoir is at \(625 \mathrm{~K}\), what are (a) the rate of energy extraction from the hot reservoir, (b) the engine's efficiency, and (c) the temperature of the ambient environment?

(a) Continue the calculation begun on page 372 in the subsection "Irreversible Heat Transfer" to derive the expression given in the text for the entropy change when equal masses \(m\) of hot and cold water, at temperatures \(T_{\mathrm{h}}\) and \(T_{c}\), respectively, are mixed: \(\Delta S=m c \ln \left[\left(T_{c}+T_{\mathrm{h}}\right)^{2} / 4 T_{\mathrm{c}} T_{\mathrm{h}}\right]\). (b) Show that the argument of the logarithm in this expression is greater than 1 for \(T_{\mathrm{h}} \neq T_{c}\), thus showing that \(\Delta S\) is positive. Hint: This is equivalent to showing that \(\left(T_{c}+T_{\mathrm{h}}\right)^{2}>4 T_{\mathrm{c}} T_{\mathrm{h}}\). Expand the left side of this inequality, subtract \(4 T_{\mathrm{c}} T_{\mathrm{h}}\) from both sides, factor the resulting left side, and you'll have your result.

A Carnot engine extracts heat from a block of mass \(m\) and specific heat \(c\) initially at temperature \(T_{\mathrm{b} 0}\) but without a heat source to maintain that temperature. The engine rejects heat to a reservoir at constant temperature \(T_{c}\). The engine is operated so its mechanical power output is proportional to the temperature difference \(T_{\mathrm{h}}-T_{\mathrm{c}}\) : $$ P=P_{0} \frac{T_{\mathrm{h}}-T_{\mathrm{c}}}{T_{\mathrm{h} 0}-T_{\mathrm{c}}} $$ where \(T_{\mathrm{h}}\) is the instantaneous temperature of the hot block and \(P_{0}\) is the initial power. (a) Find an expression for \(T_{\mathrm{h}}\) as a function of time, and (b) determine how long it takes for the engine's power output to reach zero.
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