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Chapter 19: Problem 41

By means of an electric immersion heater connected to 240 volt mains, it is required to heat 100 litres of water from \(20^{\circ} \mathrm{C}\) to \(70^{\circ} \mathrm{C}\) in 2 hours. Assuming no heat loss. the current necessary to effect this is (a) \(20.2 \mathrm{amp}\) (b) \(12.1 \mathrm{amp}\) (c) \(6 \mathrm{amp}\) (d) \(18.1 \mathrm{amp}\)

Short Answer

Expert verified
The current necessary is 12.1 amp (option b).

Step by step solution

01

Determine the Energy Required

To find the energy needed to heat the water, we use the formula \( Q = mc\Delta T \). Here, \( m \) is the mass of the water, \( c \) is the specific heat capacity of water, and \( \Delta T \) is the change in temperature. We have 100 liters of water, which is 100 kg (since the density of water is 1 kg/L). The specific heat capacity of water is \( 4.186 \text{ kJ/kg}^\circ\text{C} \), and the temperature change \( \Delta T \) is \( 70 - 20 = 50 \text{掳C} \). Therefore, \( Q = 100 \times 4.186 \times 50 = 20930 \text{ kJ} \) or \( 20930000 \text{ J} \).
02

Determine the Power Required

The power \( P \) required to deliver this energy in 2 hours can be found using the formula \( P = \frac{Q}{t} \), where \( Q \) is the energy in joules and \( t \) is the time in seconds. The time \( t \) is 2 hours, which is \( 2 \times 3600 = 7200 \) seconds. Thus, \( P = \frac{20930000}{7200} = 2906.94 \text{ W} \).
03

Determine the Current Required

Using the formula \( P = VI \), where \( V \) is the voltage (240 V) and \( I \) is the current in amperes, we can find the current. Rearranging gives \( I = \frac{P}{V} \). Substituting the known values, \( I = \frac{2906.94}{240} = 12.11 \text{ A} \).
04

Matching with Given Options

From the calculated current, \( I = 12.11 \text{ A} \), we compare with the given options. The closest match is option (b): \( 12.1 \mathrm{amp} \).

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Key Concepts

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

Specific Heat Capacity of Water
Water has a unique property known as specific heat capacity, which is the amount of energy required to raise the temperature of 1 kilogram of a substance by 1 degree Celsius. For water, this is particularly high at 4.186 kJ/kg掳C. This means water can absorb a lot of heat without a significant change in temperature. This high specific heat capacity is why water is used in heating and cooling systems. To calculate the energy needed to heat water, you multiply its mass by the specific heat capacity and the temperature change. This formula is:
  • \( Q = mc\Delta T \)
  • Where \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change.
This concept is crucial in understanding how much energy an electric immersion heater must supply to achieve the desired warmth.
Current Calculation
An electric immersion heater uses electrical energy to produce heat. To find out how much current is required to run the heater, we need to calculate the power needed first. Using the relation between power, voltage, and current is essential here. Power (P) can be described by the formula:
  • \( P = VI \)
  • Where \( V \) is the voltage, and \( I \) is the current.
You can rearrange this to find the current: \( I = \frac{P}{V} \). With all other values known, this calculation becomes straightforward. This helps in matching the current to the operational needs of your device.
Energy Required to Heat Water
To determine how much energy is required to heat a given volume of water, you not only need to consider the specific heat capacity of water but also its mass and the change in temperature you desire. Using the energy formula:
  • \( Q = mc\Delta T \)
you can see how each variable contributes to the total energy requirement. The mass of the water in kilograms and the temperature change both proportionally increase the energy demand. This calculation effectively tells you how much thermal energy needs to be supplied to achieve your heating goals.
Power Calculation
Once the energy requirement is known, the next step is to determine how fast that energy should be supplied, which is essentially the power. Power, measured in watts, tells you the rate at which energy is used or transferred. The formula \( P = \frac{Q}{t} \) helps here, where Q is the energy in joules and \( t \) is the time in seconds. For instance, heating water over a shorter period demands a higher power. Conversely, with the same energy, a longer heating time reduces the power requirement. Thus, power calculation is pivotal in planning the operational efficiency and capability of an electric immersion heater.
Voltage and Current Relationship
In any electrical device, like an electric immersion heater, understanding the relationship between voltage and current is key for safe and efficient operation. Ohm鈥檚 Law primarily governs this relationship through the formula \( V = IR \), but in terms of power, we refer to \( P = VI \). This formula tells you that for a given power demand, a higher voltage supply means a lower current requirement, and vice versa. This inverse relationship between voltage and current helps in determining the best electrical configuration needed for efficient heating. It also ensures that the electrical components are not overloaded, promoting longevity and safety of the appliance.

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Most popular questions from this chapter

ECE of \(\mathrm{Cu}\) and \(\mathrm{Ag}\) are \(7 \times 10^{-6}\) and \(1.2 \times 10^{-6} .\) A certain current deposits \(14 \mathrm{~g}\) of Cu. Amount of Ag deposited is (a) \(1.2 \mathrm{~g}\) (b) \(1.6 \mathrm{~g}\) (c) \(2.4 \mathrm{~g}\) (d) \(1.8 \mathrm{~g}\)

The electrochemical equivalent of a metal is \(3.3 \times 10^{-7} \mathrm{~kg}\) per coulomb. The mass of the metal liberated at the cathode when a \(3 \mathrm{~A}\) current is passed for 2 seconds will be: (a) \(19.8 \times 10^{-7} \mathrm{~kg}\) (b) \(9.9 \times 10^{-7} \mathrm{~kg}\) (c) \(6.6 \times 10^{-7} \mathrm{~kg}\) (d) \(1.1 \times 10^{-7} \mathrm{~kg}\)

A constant voltage is applied between the two ends of a uniform metallic wire. Some heat is developed in it. The heat developed is doubled if: (a) both the length and radius of the wire are halved (b) both the length and radius of the wire are doubled (c) the radius of the wire is doubled (d) the length of the wire is doubled

Which one of the following statements is wrong? (a) Seebeck effect is irreversible (b) Peltier effect is localized at the junction (c) Joule heating effect is independent of the direction of the flow of the current in a conductor (d) Thomson effect is similar in origin to that of Peltier effect. They are both due to non-uniform distribution of electrons in metals

Two heating coils one of fine wire and the other of thick wire made of the same material and of the same length are connected in series and in parallel. Which of the following statements is correct? (a) In series fine wire will liberate more energy while in parallel thick one will liberate more energy. (b) In series fine wire will liberate less energy while in parallel more energy. (c) Both will liberate equal energy. (d) In series thick wire will liberate more energy, while in parallel less energy.
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