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Chapter 1: Problem 26

\(1.2 \mathrm{~kg}\) of liquid water initially at \(15^{\circ} \mathrm{C}\) is to be heated to \(95^{\circ} \mathrm{C}\) in a teapot equipped with a \(1200-\mathrm{W}\) electric heating element inside. The teapot is \(0.5 \mathrm{~kg}\) and has an average specific heat of \(0.7 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). Taking the specific heat of water to be \(4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\) and disregarding any heat loss from the teapot, determine how long it will take for the water to be heated.

Short Answer

Expert verified
Based on the given information, it takes about 6 minutes to heat the water in the teapot from 15掳C to 95掳C using a 1200 W electric heating element. This is determined by calculating the heat energy required for both the water and the teapot and dividing the total heat energy by the power of the heating element.

Step by step solution

01

Calculate the heat energy required to heat the water

First, we need to determine the heat energy absorbed by the water. We can do this using the formula: \(Q_{water} = m_{water} \times c_{water} \times (\Delta T_{water})\) Where: - \(Q_{water}\) is the heat energy absorbed by the water - \(m_{water}\) is the mass of the water (\(1.2 \mathrm{~kg}\)) - \(c_{water}\) is the specific heat capacity of water (\(4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\)) - \(\Delta T_{water}\) is the temperature difference experienced by the water (80掳C) We can now plug in the values and calculate the heat energy required to heat the water: \(Q_{water} = 1.2 \mathrm{~kg} \times 4.18 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \times 80 \mathrm{~K} = 402.048 \mathrm{~kJ}\)
02

Calculate the heat energy required to heat the teapot

Similarly, we now need to determine the heat energy absorbed by the teapot using the formula: \(Q_{teapot} = m_{teapot} \times c_{teapot} \times (\Delta T_{teapot})\) Where: - \(Q_{teapot}\) is the heat energy absorbed by the teapot - \(m_{teapot}\) is the mass of the teapot (\(0.5 \mathrm{~kg}\)) - \(c_{teapot}\) is the specific heat capacity of the teapot (\(0.7 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\)) - \(\Delta T_{teapot}\) is the temperature difference experienced by the teapot (80掳C) We can now plug in the values and calculate the heat energy required to heat the teapot: \(Q_{teapot} = 0.5 \mathrm{~kg} \times 0.7 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \times 80 \mathrm{~K} = 28 \mathrm{~kJ}\)
03

Calculate the total heat energy required

To find the total heat energy required to heat the water and the teapot, we simply add the heat energy required for each component: \(Q_{total} = Q_{water} + Q_{teapot} = 402.048 \mathrm{~kJ} + 28 \mathrm{~kJ} = 430.048 \mathrm{~kJ}\)
04

Calculate the heating time

Finally, we can find the time it takes for the water to be heated by dividing the total heat energy required by the power of the heating element. Note that we also need to convert the power from watts to kilowatts by dividing it by 1000: \(P = 1200 \mathrm{~W} = 1.2 \mathrm{~kW}\) Time \(t\) is given by: \(t = \frac{Q_{total}}{P} = \frac{430.048 \mathrm{~kJ}}{1.2 \mathrm{~kW}} = 358.373 \mathrm{~s}\) The water and the teapot will be heated to the desired temperature in approximately 358.373 seconds (or about 6 minutes).

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Key Concepts

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

Specific Heat Capacity
The concept of specific heat capacity is fundamental in understanding how materials heat up. Specific heat capacity refers to the amount of heat required to change the temperature of one kilogram of a substance by one degree Celsius (or Kelvin). Different substances have different specific heat capacities due to their unique molecular structures, which affect how they store thermal energy.
For instance:
  • Water has a high specific heat capacity, approximately 4.18 \(\text{kJ/ (kg} \cdot \text{K)}\), meaning it requires more energy to increase its temperature compared to many other materials.
  • In the exercise, the teapot also has its own specific heat capacity, measured at 0.7 \(\text{kJ/ (kg} \cdot \text{K)}\), illustrating that less energy is needed to heat it compared to water.
Understanding specific heat capacity helps in calculating how much energy is needed for heating in various practical applications.
Electric Heating
Electric heating converts electrical energy into heat energy, which is then used to raise the temperature of a substance. In this exercise, the teapot is equipped with an internal electric heating element, essential for this transformation process.
Key points to consider:
  • The power of the heating element is given as 1200 Watts, indicating the rate at which it draws electrical energy to generate heat.
  • To use the element efficiently, understanding the power output in kilowatts helps in energy calculations, as seen with the conversion to 1.2 kW in the solution.
  • Electric heating is popular because it's controllable and effective, with minimal heat loss under ideal conditions.
This concept underpins many modern heating solutions, from water heaters to kitchen appliances.
Energy Calculation
Energy calculation involves determining the total amount of energy required to achieve a certain temperature change in one or more substances. This is done using formulas linked to specific heat capacities and temperature changes. In the context of the exercise, calculations were performed separately for both the water and the teapot.
Here's how it works:
  • The formula \(Q = mc\Delta T\) calculates the energy required. Here, \(m\) is mass, \(c\) is specific heat capacity, and \(\Delta T\) is the change in temperature.
  • By substituting the known values of mass, specific heat capacity, and temperature change, you obtain the energy required for each component.
  • The total energy needed is simply the sum of the energy needed by the water and the teapot.
Energy calculations allow us to predict the energy consumption for practical heating processes, optimizing resource use.
Temperature Change
Temperature change is the difference in the thermal state of a substance, which can be increased by the application of heat. In this problem, the change in temperature is significant, as water is heated from 15掳C to 95掳C, a change of 80掳C.
Considerations in calculating temperature change:
  • It determines the amount of energy required, as a direct component in the formula \(Q = mc\Delta T\).
  • The larger the temperature change, the more energy is required, especially for substances with high specific heat capacities, like water.
  • This concept is crucial in many heating-related calculations, ensuring efficiency and accuracy in predicting energy requirements.
Understanding temperature change helps define the parameters for effective and energy-efficient heating solutions.

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Most popular questions from this chapter

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) \(12,000 \mathrm{~kJ}\)

A transistor with a height of \(0.4 \mathrm{~cm}\) and a diameter of \(0.6 \mathrm{~cm}\) is mounted on a circuit board. The transistor is cooled by air flowing over it with an average heat transfer coefficient of \(30 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the air temperature is \(55^{\circ} \mathrm{C}\) and the transistor case temperature is not to exceed \(70^{\circ} \mathrm{C}\), determine the amount of power this transistor can dissipate safely. Disregard any heat transfer from the transistor base.

One way of measuring the thermal conductivity of a material is to sandwich an electric thermofoil heater between two identical rectangular samples of the material and to heavily insulate the four outer edges, as shown in the figure. Thermocouples attached to the inner and outer surfaces of the samples record the temperatures. During an experiment, two \(0.5-\mathrm{cm}\) thick samples \(10 \mathrm{~cm} \times\) \(10 \mathrm{~cm}\) in size are used. When steady operation is reached, the heater is observed to draw \(25 \mathrm{~W}\) of electric power, and the temperature of each sample is observed to drop from \(82^{\circ} \mathrm{C}\) at the inner surface to \(74^{\circ} \mathrm{C}\) at the outer surface. Determine the thermal conductivity of the material at the average temperature.

The inner and outer surfaces of a \(25-\mathrm{cm}\)-thick wall in summer are at \(27^{\circ} \mathrm{C}\) and \(44^{\circ} \mathrm{C}\), respectively. The outer surface of the wall exchanges heat by radiation with surrounding surfaces at \(40^{\circ} \mathrm{C}\), and convection with ambient air also at \(40^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(8 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Solar radiation is incident on the surface at a rate of \(150 \mathrm{~W} / \mathrm{m}^{2}\). If both the emissivity and the solar absorptivity of the outer surface are \(0.8\), determine the effective thermal conductivity of the wall.

A thin metal plate is insulated on the back and exposed to solar radiation on the front surface. The exposed surface of the plate has an absorptivity of \(0.7\) for solar radiation. If solar radiation is incident on the plate at a rate of \(550 \mathrm{~W} / \mathrm{m}^{2}\) and the surrounding air temperature is \(10^{\circ} \mathrm{C}\), determine the surface temperature of the plate when the heat loss by convection equals the solar energy absorbed by the plate. Take the convection heat transfer coefficient to be \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), and disregard any heat loss by radiation.
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