/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 12 A \(5-\mathrm{kg}\) block of cop... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 12

A \(5-\mathrm{kg}\) block of copper at \(300^{\circ} \mathrm{C}\) is submerged in 20 liters of water at \(0^{\circ} \mathrm{C}\) contained in an insulated tank. Estimate the final equilibrium temperature. SOLUTION Conservation of energy requires that the energy lost by the copper block is gained by the water. This is expressed as $$ m_{c}\left(C_{p}\right)_{c}(\Delta T)_{c}=m_{w}\left(C_{p}\right)_{w}(\Delta T)_{w} $$ Using average values of \(C_{p}\) from Table B.4, this becomes $$ (5)(0.39)\left(300-T_{2}\right)=(0.02)(1000)(4.18)\left(T_{2}-0\right) \quad \therefore T_{2}=6.84^{\circ} \mathrm{C} $$

Short Answer

Expert verified
The final equilibrium temperature is \( 6.84^{\text{掳C}} \).

Step by step solution

01

Understand the Problem

A block of copper at a higher temperature is submerged in water at a lower temperature in an insulated tank. The goal is to find the final equilibrium temperature where both have the same temperature.
02

Conservation of Energy

Apply the principle of conservation of energy. The energy lost by the copper block must equal the energy gained by the water. This can be expressed as: \[ m_c C_p_c (\triangle T)_c = m_w C_p_w (\triangle T)_w \] where \( m_c \) and \( m_w \) are the masses, \( C_p_c \) and \( C_p_w \) are the specific heat capacities, and \( \triangle T \) refers to the change in temperature for copper and water, respectively.
03

Substitute Known Values

From the problem, \( m_c = 5 \text{ kg} \), \( C_p_c = 0.39 \text{ kJ/kg路掳C} \), \( \triangle T_c = 300 - T_2 \), \( m_w = 20 \text{ liters} = 20 \text{ kg (since 1 liter of water = 1 kg)} \), \( C_p_w = 4.18 \text{ kJ/kg路掳C} \), and \( \triangle T_w = T_2 - 0 \). Substitute these values into the conservation equation: \[ (5)(0.39)(300 - T_2) = (20)(4.18)(T_2 - 0) \]
04

Solve for \( T_2 \)

Simplify and solve for \( T_2 \): \[ (5)(0.39)(300 - T_2) = (20)(4.18)(T_2) \] \[ 1.95 (300 - T_2) = 83.6 T_2 \] \[ 585 - 1.95 T_2 = 83.6 T_2 \] \[ 585 = 85.55 T_2 \] \[ T_2 = \frac{585}{85.55} \] \[ T_2 = 6.84^{\text{掳C}} \]

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Key Concepts

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

Conservation of Energy
Conservation of energy is a fundamental principle in thermodynamics. It states that energy cannot be created or destroyed, only transferred or transformed. In our exercise, the energy lost by the hot copper block must equal the energy gained by the cold water. This ensures the total energy in the system remains constant.This relationship is mathematically represented by: \[ m_c C_p_c (\triangle T)_c = m_w C_p_w (\triangle T)_w \] Where:
  • \( m_c \) and \( m_w \) are the masses of copper and water, respectively.
  • \( C_p_c \) and \( C_p_w \) are the specific heat capacities of copper and water.
  • \( \triangle T \) denotes the change in temperature for copper and water.
In solving the problem, we set up the energy balance equation and use known values to find the unknown final temperature. This showcases how energy conservation principles help us predict outcomes in thermodynamic systems.
Specific Heat Capacity
Specific heat capacity is a property that describes how much heat energy is required to raise the temperature of a substance by one degree Celsius. It varies from material to material.In our problem:
  • Copper has a specific heat capacity of 0.39 kJ/kg路掳C.
  • Water has a specific heat capacity of 4.18 kJ/kg路掳C.
The difference in these values indicates that water requires more energy to change temperature compared to copper. This concept is critical in determining how much heat will be transferred between the two substances. When solving the exercise, we utilized the specific heat capacities to calculate the equilibrium temperature. This ensures that the heat lost by the copper (due to its higher initial temperature) equals the heat gained by the water (starting at a lower temperature).
Thermodynamic Equilibrium
Thermodynamic equilibrium is achieved when two or more substances in contact with each other reach the same temperature, resulting in no net heat flow between them.In our exercise, this is the final state we calculated, where the copper and water attain a common temperature of 6.84掳C.To find this equilibrium temperature, we used the energy balance from the conservation of energy principle: \[ m_c C_p_c (\triangle T)_c = m_w C_p_w (\triangle T)_w \]By solving this equation, we ensured both substances have balanced their heat energy. This maintained the overall energy in the insulated system, leading to thermodynamic equilibrium.Understanding this concept helps in problems involving heat transfer, phase changes, and various other thermodynamic processes.

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Most popular questions from this chapter

Estimate the constant-pressure specific heat and the constant-volume specific heat for R134a at 30 psia and \(100^{\circ} \mathrm{F}\). SOLUTION We write the derivatives in finite-difference form and, using values on either side of \(100^{\circ} \mathrm{F}\) for greatest accuracy, we find $$ \begin{aligned} &C_{p} \equiv \frac{\Delta h}{\Delta T}=\frac{126.39-117.63}{120-80}=0.219 \mathrm{Btu} / \mathrm{lbm}-{ }^{\circ} \mathrm{F} \\ &C_{v} \equiv \frac{\Delta u}{\Delta T}=\frac{115.47-107.59}{120-80}=0.197 \mathrm{Btu} / \mathrm{lbm}-{ }^{\circ} \mathrm{F} \end{aligned} $$

A \(1500-\mathrm{kg}\) automobile traveling at \(30 \mathrm{~m} / \mathrm{s}\) is brought to rest by impacting a shock absorber composed of a piston with small holes that moves in a cylinder containing water. How much heat must be removed from the water to return it to its original temperature? As the piston moves through the water, work is done due to the force of impact moving with the piston. The work that is done is equal to the kinetic energy change; that is, $$ W=\frac{1}{2} m V^{2}=\left(\frac{1}{2}\right)(1500)(30)^{2}=675000 \mathrm{~J} $$ The first law for a cycle requires that this amount of heat must be transferred from the water to return it to its original temperature; hence, \(Q=675 \mathrm{~kJ}\).

Steam with a mass flux of \(600 \mathrm{lbm} / \mathrm{min}\) exits a turbine as saturated steam at 2 psia and passes through a condenser (a heat exchanger). What mass flux of cooling water is needed if the steam is to exit the condenser as saturated liquid and the cooling water is allowed a \(15^{\circ} \mathrm{F}\) temperature rise? SOLUTION The energy equations (4.75) are applicable to this situation. The heat transfer rate for the steam is, assuming no pressure drop through the condenser, $$ \dot{Q}_{s}=\dot{m}_{s}\left(h_{s 2}-h_{s 1}\right)=(600)(94.02-1116.1)=-613.200 \mathrm{Btu} / \mathrm{min} $$ This energy is gained by the water. Hence, $$ \dot{Q}_{w}=\dot{m}_{w}\left(h_{w 2}-h_{w 1}\right)=\dot{m}_{w} C_{p}\left(T_{w 2}-T_{w 1}\right) \quad \frac{613,200}{60}=\dot{m}_{w}(1.00)(15) \quad \dot{m}_{w}=681 \mathrm{lbm} / \mathrm{s} $$

Calculate the change in enthalpy of air which is heated from \(300 \mathrm{~K}\) to \(700 \mathrm{~K}\) if (a) \(C_{p}=1.006 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\). (b) Table B.5 is used. (c) The gas tables are used. (d) Compare the calculations of \((a)\) and \((b)\) with \((c)\). SOLUTION (a) Assuming the constant specific heat, we find that $$ \Delta h=C_{p}\left(T_{2}-T_{1}\right)=(1.006)(700-300)=402 \mathrm{~kJ} / \mathrm{kg} $$ (b) If \(C_{p}\) depends on temperature, we must integrate as follows: $$ \Delta h=\frac{1}{M} \int_{T_{1}}^{T_{2}} C_{p} d T=\frac{1}{28.97} \int_{300}^{700}\left(28.11+0.00197 T+0.48 \times 10^{-5} T^{2}-1.97 \times 10^{-9} T^{3}\right) d T=415 \mathrm{~kJ} / \mathrm{kg} $$ (c) Using Table E.1, we find \(\Delta h=h_{2}-h_{1}=713.27-300.19=413 \mathrm{~kJ} / \mathrm{kg}\). (d) The assumption of constant specific heat results in an error of \(-2.6\) percent; the expression for \(C_{p}\) produces an error of \(+0.48\) percent. All three methods are acceptable assuming the gas tables are most accurate.

4.29 Refrigerant R134a enters the compressor of the simple refrigeration cycle of Fig. \(4.19\) as a saturated vapor at \(120 \mathrm{kPa}\), enters the condenser at \(800 \mathrm{kPa}\) and \(50^{\circ} \mathrm{C}\), exits the condenser as a saturated liquid, and exits the evaporator as a saturated vapor. Calculate \((a)\) the lowest cycle temperature, \((b)\) the heat transferred from the condenser, \((c)\) the cooling load \(\dot{Q}_{E},(d)\) the necessary compressor horsepower, and \((e)\) the coefficient of performance for this refrigerator. The refrigerant mass flow rate is \(0.2 \mathrm{~kg} / \mathrm{s}\). The enthalpies are of primary interest into and out of the four control volumes. The IRC Fluid Property Calculator will be used to obtain the enthalpies, but, if, for some reason, it is not available, Appendix D can be used (the IRC Fluid Property Calculator gives slightly different values than does Table D). They are listed in the following: State 1 (saturated vapor): \(\quad P_{1}=P_{4}=120 \mathrm{kPa}, x=1 . \therefore h_{1}=237 \mathrm{~kJ} / \mathrm{kg}\) State 2 (superheated vapor): \(\quad P_{2}=800 \mathrm{kPa}, T_{2}=50^{\circ} \mathrm{C} . \therefore h_{2}=287 \mathrm{~kJ} / \mathrm{kg}\) State 3 (saturated liquid): \(\quad P_{3}=P_{2}=800 \mathrm{kPa}, x_{3}=0 . \therefore h_{3}=95.5 \mathrm{~kJ} / \mathrm{kg}\) State 4 (quality region): \(\quad P_{4}=P_{1}=120 \mathrm{kPa}, h_{4}=h_{3}=95.5 \mathrm{~kJ} / \mathrm{kg}\) Using the above enthalpies, the quantities of interest can be determined. (a) The lowest cycle temperature is \(T_{4}\) (refer to Fig. 4.19). The IRC Fluid Property Calculator, with \(P_{4}=120 \mathrm{kPa}\) and \(h_{4}=95.5 \mathrm{~kJ} / \mathrm{kg}\), provides \(T_{4}=-22.3^{\circ} \mathrm{C}\). (b) The heat transferred from the condenser (a heat exchanger) is $$ \dot{Q}_{C}=\dot{m}\left(h_{2}-h_{3}\right)=0.2 \mathrm{~kg} / \mathrm{s} \times(287-95.5) \text { But } / \mathrm{lbm}=38.3 \mathrm{~kJ} / \mathrm{s} $$ (c) The cooling load provided by the evaporator (a heat exchanger) is $$ \dot{Q}_{E}=\dot{m}\left(h_{1}-h_{4}\right)=0.2(237-95.5)=28.3 \mathrm{~kJ} / \mathrm{s} $$ (d) The compressor work requirement is $$ \dot{W}_{\text {Comp }}=\dot{m}\left(h_{2}-h_{1}\right)=0.2 \times(287-237)=10 \mathrm{~kW} \quad \text { or } \quad 13.4 \mathrm{hp} $$ (e) The coefficient of performance for this ideal cycle is $$ \mathrm{COP}=\frac{\text { desired effect }}{\text { required input }}=\frac{\dot{Q}_{E}}{\dot{W}_{\text {Comp }}}=\frac{28.3}{10}=2.83 $$ If the desired effect was to heat a space, \(\dot{Q}_{C}\) would be used as the numerator in COP.
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