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Chapter 1: Problem 150

Water enters a pipe at \(20^{\circ} \mathrm{C}\) at a rate of \(0.50 \mathrm{~kg} / \mathrm{s}\) and is heated to \(60^{\circ} \mathrm{C}\). The rate of heat transfer to the water is (a) \(20 \mathrm{~kW}\) (b) \(42 \mathrm{~kW}\) (c) \(84 \mathrm{~kW}\) (d) \(126 \mathrm{~kW}\) (e) \(334 \mathrm{~kW}\)

Short Answer

Expert verified
Answer: The rate of heat transfer to the water is \(84 \mathrm{~kW}\).

Step by step solution

01

Identify relevant information and equations

We are given the initial temperature (Ti) of water as \(20^{\circ} \mathrm{C}\), the final temperature (Tf) as \(60^{\circ} \mathrm{C}\), the mass flow rate (m) as \(0.50 \mathrm{~kg}/\mathrm{s}\), and the specific heat capacity (c) of water as \(4200 \mathrm{~J}/(\mathrm{kg}\cdot\mathrm{K})\). The rate of heat transfer (Q) can be calculated using the following equation: \(Q = cm(Tf - Ti)\) Where Q is the rate of heat transfer, c is the specific heat capacity, m is the mass flow rate, Tf and Ti are the final and initial temperatures, respectively.
02

Calculate the temperature difference

Calculate the difference between the final and initial temperature, which is the change in temperature: \(\Delta T = Tf - Ti = 60^{\circ} \mathrm{C} - 20^{\circ} \mathrm{C} = 40^{\circ} \mathrm{C}\)
03

Calculate the rate of heat transfer

Now, substitute the given values into the equation and calculate the rate of heat transfer: \(Q = (4200 \mathrm{~J}/(\mathrm{kg}\cdot\mathrm{K})) \times (0.50 \mathrm{~kg}/\mathrm{s}) \times 40\mathrm{K}\) \(Q = 84000 \mathrm{~J}/\mathrm{s} = 84 \mathrm{~kW}\) Thus, the rate of heat transfer to the water is option (c) \(84 \mathrm{~kW}\).

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Key Concepts

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

Specific Heat Capacity
Specific heat capacity is a vital topic in understanding how substances respond to heat. It tells us how much energy is needed to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). For water, this value is particularly high at 4200 J/(kg路K).
This means water requires more heat energy compared to most other substances to achieve the same temperature increase.
This high specific heat capacity plays a crucial role in many applications, from cooking to industrial processes, where controlling heat is vital.
  • Energy Absorption: Water can absorb a lot of heat without a significant temperature rise, making it ideal for heating applications.
  • Protection Against Temperature Fluctuations: This property enables water to stabilize temperatures in environments, acting as a buffer against rapid changes.
In solving our problem, understanding the specific heat capacity of water helps us predict how much heat is needed to achieve the desired temperature change.
Temperature Difference
Temperature difference, often denoted by \(\Delta T\), is the change between two temperatures. It鈥檚 essential in calculating how much energy is needed for heating or cooling a material.

In our exercise, the temperature difference is calculated by subtracting the initial temperature of the water from the final temperature: 60掳C - 20掳C = \(40\, ^{\circ} \mathrm{C}\) or in Kelvin as well, since the size of degrees is the same, making it convenient.
  • Direct Relation: The larger the temperature difference, the more energy is typically required to change the temperature of the substance.
  • Practical Importance: Knowing the temperature difference helps in various practical applications, such as designing heating systems or evaluating the efficiency of thermal processes.
This concept helps simplify complex heat calculations by emphasizing the energy change instead of the specific initial and final temperatures.
Mass Flow Rate
Mass flow rate is the quantity of mass passing through a given surface per unit time. It is a crucial factor in heat transfer calculations as it determines how much substance is being heated or cooled in a given timeframe.
In the context of our problem, where water's mass flow rate is 0.50 kg/s, this means that each second, 0.50 kg of water flows through the heating system.
This concept allows us to figure out how much heat is needed per second, given the specific heat capacity and temperature difference.
  • Heat Transfer Influence: A higher mass flow rate can mean more energy is required to achieve the same temperature change, as more material is being processed.
  • System Design: Engineers must consider mass flow rates in designing and analyzing thermal systems to ensure efficient energy usage.
Understanding mass flow rate is crucial for accurate energy usage predictions, ensuring systems meet desired thermal outputs efficiently.

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Most popular questions from this chapter

Consider a sealed 20-cm-high electronic box whose base dimensions are \(50 \mathrm{~cm} \times 50 \mathrm{~cm}\) placed in a vacuum chamber. The emissivity of the outer surface of the box is \(0.95\). If the electronic components in the box dissipate a total of \(120 \mathrm{~W}\) of power and the outer surface temperature of the box is not to exceed \(55^{\circ} \mathrm{C}\), determine the temperature at which the surrounding surfaces must be kept if this box is to be cooled by radiation alone. Assume the heat transfer from the bottom surface of the box to the stand to be negligible.

An electronic package in the shape of a sphere with an outer diameter of \(100 \mathrm{~mm}\) is placed in a large laboratory room. The surface emissivity of the package can assume three different values \((0.2,0.25\), and \(0.3)\). The walls of the room are maintained at a constant temperature of \(77 \mathrm{~K}\). The electronics in this package can only operate in the surface temperature range of \(40^{\circ} \mathrm{C} \leq T_{s} \leq 85^{\circ} \mathrm{C}\). Determine the range of power dissipation \((\dot{W})\) for the electronic package over this temperature range for the three surface emissivity values \((\varepsilon)\). Plot the results in terms of \(\dot{W}(\mathrm{~W})\) vs. \(T_{s}\left({ }^{\circ} \mathrm{C}\right)\) for the three different values of emissivity over a surface temperature range of 40 to \(85^{\circ} \mathrm{C}\) with temperature increments of \(5^{\circ} \mathrm{C}\) (total of 10 data points for each \(\varepsilon\) value). Provide a computer generated graph for the display of your results and tabulate the data used for the graph. Comment on the results obtained.

Consider a person standing in a room at \(18^{\circ} \mathrm{C}\). Determine the total rate of heat transfer from this person if the exposed surface area and the skin temperature of the person are \(1.7 \mathrm{~m}^{2}\) and \(32^{\circ} \mathrm{C}\), respectively, and the convection heat transfer coefficient is \(5 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Take the emissivity of the skin and the clothes to be \(0.9\), and assume the temperature of the inner surfaces of the room to be the same as the air temperature.

Can all three modes of heat transfer occur simultaneously (in parallel) in a medium?

A 2-kW electric resistance heater in a room is turned on and kept on for 50 minutes. The amount of energy transferred to the room by the heater is (a) \(2 \mathrm{~kJ}\) (b) \(100 \mathrm{~kJ}\) (c) \(6000 \mathrm{~kJ}\) (d) \(7200 \mathrm{~kJ}\) (e) \(12,000 \mathrm{~kJ}\)
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