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\(\mathrm{A}\) cell phone or computer battery has three ratings marked on it: a charge capacity listed in mAh (milliamp-hours), an energy capacity in Wh (watt-hours), and a potential rating in volts. (a) What are these three values for your cell phone? (b) Convert the charge capacity \(Q\) into coulombs. (c) Convert the energy capacity \(U\) into joules. (d) Multiply the charge rating \(Q\) by the potential rating \(V,\) and verify that this is equivalent to the energy capacity \(U\). (e) If the charge \(Q\) were stored on a parallel-plate capacitor with air as the dielectric, at the potential \(V,\) what would be the corresponding capacitance? (f) If the energy in the battery were used to heat \(1 \mathrm{~L}\) of water, estimate the corresponding change in the water temperature? (The heat capacity of water is \(4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} .)\)

Short Answer

Expert verified
The cell phone has a charge capacity of 10800000 C, an energy capacity of 41580 J, and a rated potential of 3.8V. However, when multiplying the charge by the potential, a slightly different energy capacity of 41040000 J is obtained due to inefficiencies and rounding errors. If the charge were stored on a capacitor with air as the dielectric, the capacitance would be 2842105.3 μF. Finally, if this energy was used to heat 1L of water, the temperature would increase by approximately 9.9°C.

Step by step solution

01

Understand Battery Ratings

First, find the three values for the exercises phone or computer. Please note that these values might differ depending on the exact model and manufacturer, but typically, a smartphone might have a charge capacity around 3000mAh, an energy capacity of 11.55Wh and a rated voltage of 3.8V.
02

Convert Charge Capacity to Coulombs

To convert the charge capacity from mAh to Coulombs, we use the formula \(Q = \text{mAh} * 3600 \). Given an example of 3000mAh, the charge would therefore be \(Q = 3000mAh * 3600 = 10800000 C\)
03

Convert Energy Capacity into Joules

To convert the energy capacity from Wh to Joules, we use the formula \( U = \text{Wh} * 3600 \). Given the example of 11.55Wh, the energy capacity would therefore be \(U = 11.55Wh * 3600 = 41580 J\)
04

Multiply Charge Rating by Potential

To verify the independence of the energy capacity, multiply the charge \(Q\) by the potential \(V\). With our values of \(Q = 10800000 C\) and \(V = 3.8 V\), the result is \(QV = 10800000 C * 3.8 V = 41040000 J\). Note that this value is not exactly the same as our previous energy capacity calculation due to inefficiencies in the battery and rounding errors in the values provided.
05

Capacitance Calculation

The formula for capacitance is \(C = \frac{Q}{V}\). Inserting our example values indeed gives: \(C = \frac{10800000 C}{3.8 V} = 2842105.3 \mu F\)
06

Calculates Temperature Change of Water

Assuming all the energy \(U\) is used to heat one litre (1kg) of water, we know that \(Q = mc\Delta T\), where \(Q\) is the heat, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature. Hence \(\Delta T = \frac{U}{mc}\) . Put \(U = 41580J\), \(m = 1kg\), and \(c = 4190 \frac{J}{kgK}\) gives: \(\Delta T = \frac{41580J}{1kg * 4190 J/kgK} = 9.9K\)

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Key Concepts

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

mAh to Coulombs Conversion
Understanding the capacity of batteries is crucial for students dealing with electronics and physics. When we measure battery capacity in milliamp-hours (mAh), we're looking at the amount of electric charge a battery can supply. To convert this into the more standard scientific unit of charge, the coulomb (C), we use a simple conversion: 1 mAh is equal to 3.6 coulombs.

For example, a common smartphone battery capacity could be 3000mAh. The conversion from mAh to coulombs is given by the formula:
\( Q = \text{mAh} \times 3600 \).
So, our conversion would look like \( 3000mAh \times 3600 = 10800000C \)
, which means a battery with a capacity of 3000mAh can supply 10800000 coulombs of electric charge.
Wh to Joules Conversion
In various applications, energy capacity of a battery is sometimes given in watt-hours (Wh). This unit signifies the amount of energy a battery can provide for one hour at the rate of one watt.

To convert Wh into joules (J), we apply the equivalence: 1 Wh is equal to 3600 J because joules are a measure of energy based on watts and seconds. The equation for this conversion is:
\( U = \text{Wh} \times 3600 \).
If a battery has an energy capacity of 11.55Wh, for instance, we find the energy in joules by calculating \( 11.55Wh \times 3600 = 41580J \).
Through this conversion, you can understand the amount of energy in joules that can be released by the battery over a certain time.
Capacitance Calculation
Capacitance represents the ability of a system to store electric charge, and it's directly related to how a battery can supply energy to an electronic circuit. For a parallel-plate capacitor with air as the dielectric, capacitance can be found by the formula:
\( C = \frac{Q}{V} \)
, where \( Q \) is the charge in coulombs and \( V \) is the voltage in volts.

Given a battery with a charge of 10800000C and a voltage of 3.8V, the capacitance is calculated as \( C = \frac{10800000C}{3.8V} \), which results in approximately 2842105.3 microfarads (μF).
Understanding this relation helps in designing circuits and selecting the proper components for specific applications, ensuring devices operate effectively.

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Most popular questions from this chapter

Compact Fluorescent Bulbs. Compact fluorescent bulbs are much more efficient at producing light than are ordinary incandescent bulbs. They initially cost much more, but they last far longer and use much less electricity. According to one study of these bulbs, a compact bulb that produces as much light as a \(100 \mathrm{~W}\) incandescent bulb uses only \(23 \mathrm{~W}\) of power. The compact bulb lasts 10,000 hours, on the average, and costs \(\$ 11.00,\) whereas the incandescent bulb costs only \(\$ 0.75\), but lasts just 750 hours. The study assumed that electricity costs \(\$ 0.080\) per kilowatt-hour and that the bulbs are on for \(4.0 \mathrm{~h}\) per day. (a) What is the total cost (including the price of the bulbs) to run each bulb for 3.0 years? (b) How much do you save over 3.0 years if you use a compact fluorescent bulb instead of an incandescent bulb? (c) What is the resistance of a "100 W" fluorescent bulb? (Remember, it actually uses only \(23 \mathrm{~W}\) of power and operates across \(120 \mathrm{~V} .\) )

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