/*! 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 13 A small electric immersion heate... [FREE SOLUTION] | 91Ó°ÊÓ

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A small electric immersion heater is used to heat \(100 \mathrm{~g}\) of water for a cup of instant coffee. The heater is labeled "200 watts," so it converts electrical energy to thermal energy that is transferred to the water at this rate. Calculate the time required to bring the water from \(23^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) ignoring any thermal energy that transfers out of the cup.

Short Answer

Expert verified
The time required to heat the water is approximately 26.86 minutes.

Step by step solution

01

Convert mass to kilograms

First, convert the mass of water from grams to kilograms. Since there are 1000 grams in a kilogram, the mass of water is \[ m = \frac{100}{1000} = 0.1 \text{ kg} \]
02

Find the specific heat capacity of water

The specific heat capacity (\(c\)) of water is a constant and is equal to \[ c = 4186 \text{ J/(kg}^\text{°}\text{C)} \]
03

Calculate the temperature change

Find the change in temperature by subtracting the initial temperature from the final temperature: \[ \triangle T = 100^\text{°}\text{C} - 23^\text{°}\text{C} = 77^\text{°}\text{C} \]
04

Calculate the thermal energy required

Use the formula \[ Q = m \times c \times \triangle T \]to find the thermal energy required: \[ Q = 0.1 \text{ kg} \times 4186 \text{ J/(kg}^\text{°}\text{C)} \times 77^\text{°}\text{C} = 322,302 \text{ J} \]
05

Find the power converted by the heater

The power of the heater is given as 200 watts, which means it converts 200 joules of electrical energy to thermal energy per second: \[ P = 200 \text{ W} \]
06

Calculate the time required to heat the water

Use the formula \[ t = \frac{Q}{P} \]to calculate the time required: \[ t = \frac{322,302 \text{ J}}{200 \text{ J/s}} = 1611.51 \text{ s} \]Convert the time to minutes by dividing by 60: \[ t = \frac{1611.51 \text{ s}}{60 \text{ s/min}} \text{ ≈ 26.86 minutes} \]

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Key Concepts

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

Thermal Energy Transfer
Thermal energy transfer is the process where thermal energy moves from one object to another. In our exercise, the electric immersion heater transfers thermal energy to the water. The key here is understanding that energy is being transferred from the heater (electrical energy) to the water (thermal energy). This warm energy causes the water molecules to move faster, leading to an increase in temperature. The unit of measurement for energy is Joules (J), which is an essential concept to grasp. Whenever we are calculating how much energy is transferred, this is often done using the formula: \[Q = m \times c \times \triangle T \] where \(Q\) is the thermal energy, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\triangle T\) is the change in temperature. This formula is the cornerstone of many thermal energy transfer calculations.
Specific Heat Capacity
Specific heat capacity is a property of a material that tells us how much energy it takes to raise the temperature of 1 kg of the material by 1 degree Celsius (\(^\text{o}\text{C}\)). In the context of our exercise, the specific heat capacity of water is crucial because it informs us how many Joules are needed to increase the temperature of water. For water, this value is quite high at 4186 J/(kg\(^\text{o}\text{C}\)). This high value means that water requires a lot of energy to change its temperature, which is why it is effective for heating and cooling purposes. When performing calculations involving specific heat capacity, always ensure you're using the correct value for the material in question.
Power and Energy Conversion
Power is defined as the rate at which energy is transferred or converted. In our example, the immersion heater converts electrical energy into thermal energy at a rate of 200 watts. A watt is a unit of power and when we say '200 watts,' it means the heater converts 200 Joules of energy every second. So, power can be calculated using: \[P = \frac{Q}{t}\] where \(P\) is the power, \(Q\) is the energy, and \(t\) is the time over which the energy conversion occurs. Understanding this concept helps us link the amount of energy required to heat the water with the time it will take, given the power rating of the heater.
Temperature Change Calculation
The calculation of temperature change is a straightforward yet vital step. In the exercise, we determine the initial and final temperatures of the water to find the temperature change. This is done using: \[\triangle T = T_{\text{final}} - T_{\text{initial}}\] For our example, this is: \[\triangle T = 100^\text{o} \text{C} - 23^\text{o} \text{C} = 77^\text{o} \text{C}\] This means that the water's temperature needed to increase by 77 degrees Celsius to reach boiling point. Such a calculation is necessary whenever you're dealing with heating or cooling processes, as it determines the amount of energy required for the process.
Time Calculation in Physical Systems
The final step is to calculate how long it will take to heat the water using the immersion heater. Knowing the total energy needed and the power of the heater, we use the formula: \[t = \frac{Q}{P}\] where \(t\) is the time, \(Q\) is the thermal energy, and \(P\) is the power. From our exercise: \[t = \frac{322,302 \text{ J}}{200 \text{ J/s}} = 1611.51 \text{ s}\] Converting seconds to minutes, we get: \[t = \frac{1611.51 \text{ s}}{60 \text{ s/min}} ≈ 26.86 \text{ minutes}\] This calculation informs us how long it will take under ideal conditions, and is invaluable in practical applications where timing is crucial.

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Most popular questions from this chapter

A power of \(0.400 \mathrm{hp}\) is required for \(2.00\) min to drill a hole in a \(1.60\) lb copper block. (a) If the full power is the rate at which thermal energy is generated, how much is generated in Btu? (b) What is the rise in temperature of the copper if the copper absorbs \(75.0 \%\) of this energy? (Use the energy conversion \(1 \mathrm{ft} \cdot \mathrm{lb}=\) \(\left.1.285 \times 10^{-3} \mathrm{Btu} .\right)\)

A cylindrical copper rod of length \(1.2 \mathrm{~m}\) and cross-sectional area \(4.8 \mathrm{~cm}^{2}\) is insulated to prevent thermal energy from being transferred through its surface. The ends are maintained at a temperature difference of \(100^{\circ} \mathrm{C}\) by having one end in a water-ice mixture and the other in boiling water and steam. (a) Find the rate at which thermal energy is conducted along the rod. (b) Find the rate at which ice melts at the cold end.

A person makes a quantity of iced tea by mixing 500 \(\mathrm{g}\) of hot tea (essentially water) with an equal mass of ice at its melting point. If the initial hot tea is at a temperature of (a) \(90^{\circ} \mathrm{C}\) and (b) \(70^{\circ} \mathrm{C}\), what are the temperature and mass of the remaining ice when the tea and ice reach a common temperature? Neglect energy transfers with the environment.

what temperature do the following pairs of scales read the same, if ever: (a) Fahrenheit and Celsius (verify the listing in Table \(19-1\) ), (b) Fahrenheit and Kelvin, and (c) Celsius and Kelvin?

(a) Calculate the rate at which body heat is conducted through the clothing of a skier in a steady-state process, given the following data: the body surface area is \(1.8 \mathrm{~m}^{2}\) and the clothing is \(1.0 \mathrm{~cm}\) thick; the skin surface temperature is \(33^{\circ} \mathrm{C}\) and the outer surface of the clothing is at \(1.0^{\circ} \mathrm{C} ;\) the thermal conductivity of the clothing is \(0.040 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). (b) How would the answer to (a) change if, after a fall, the skier's clothes became soaked with water of thermal conductivity \(0.60 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K} ?\)

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