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Ch. 19 - Heat and the First Law of Thermodynamics
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
All textbooksGiancoli Douglas 5th editionCh. 19 - Heat and the First Law of ThermodynamicsProblem 11
Chapter 19, Problem 11

(a) How long does it take a 750-W coffeepot to bring to a boil 0.75 L of water at sea level initially at 11°C? Assume that the part of the pot which is heated with the water is made of 250 g of aluminum, and that no water boils away.
(b) For how long could this amount of energy run a 60-W lightbulb?

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Step 1: Calculate the total energy required to heat the water. Use the formula for heat transfer: Q = mcΔT, where m is the mass of the water, c is the specific heat capacity of water (approximately 4186 J/(kg·°C)), and ΔT is the temperature change. Convert the volume of water (0.75 L) to mass (in kg) using the density of water (1 kg/L).
Step 2: Calculate the energy required to heat the aluminum part of the pot. Use the same formula Q = mcΔT, but this time for aluminum. The mass of the aluminum is 250 g (convert to kg), and the specific heat capacity of aluminum is approximately 900 J/(kg·°C). Add this energy to the energy required for the water to get the total energy.
Step 3: Determine the time required to supply this energy using the coffeepot's power. Use the formula P = Q/t, where P is the power of the coffeepot (750 W), Q is the total energy calculated in the previous steps, and t is the time. Rearrange to solve for t: t = Q/P.
Step 4: For part (b), calculate the total time a 60-W lightbulb could run using the same amount of energy. Use the formula P = Q/t, but this time rearrange to solve for t: t = Q/P, where P is the power of the lightbulb (60 W).
Step 5: Ensure all units are consistent throughout the calculations (e.g., mass in kg, temperature in °C, power in watts, and energy in joules). Perform the calculations step by step to find the time for both the coffeepot and the lightbulb.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Specific Heat Capacity

Specific heat capacity is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. For water, this value is approximately 4.18 J/g°C. In this problem, it is essential to calculate the energy needed to heat the water from its initial temperature to its boiling point, which involves using the specific heat capacity of water.
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Energy Transfer and Power

Power is the rate at which energy is transferred or converted, measured in watts (W), where 1 W = 1 J/s. The coffeepot's power rating of 750 W indicates how much energy it can provide per second. To find out how long it takes to heat the water, one must calculate the total energy required and then divide it by the power of the coffeepot.
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Energy Conservation

The principle of energy conservation states that energy cannot be created or destroyed, only transformed from one form to another. In this scenario, the energy used by the coffeepot to heat the water can be calculated and then compared to the energy consumption of a 60-W lightbulb to determine how long the same amount of energy could power the bulb.
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Related Practice
Textbook Question

A 215-g sample of a substance is heated to 330°C and then plunged into a 105-g aluminum calorimeter cup containing 185 g of water and a 17-g glass thermometer at 10.5°C. The final temperature is 35.0°C. What is the specific heat of the substance? (Assume no water boils away.)

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Textbook Question

What is the specific heat of a metal substance if 165 kJ of heat is needed to raise 4.1 kg of the metal from 18.0°C to 37.2°C?

Textbook Question

The heat capacity, C, of an object is defined as the amount of heat needed to raise its temperature by 1 °C. Thus, to raise the temperature by ∆T requires heat Q given by Q = C∆T. What is the heat capacity of 38 kg of water?

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Textbook Question

An average active person consumes about 2500 Cal a day.

(a) What is this in joules?

(b) What is this in kilowatt-hours?

(c) If your power company charges about per kilowatt-hour, how much would your energy cost per day if you bought it from the power company? Could you feed yourself on this much money per day?

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Textbook Question

(II) A cube of ice is taken from the freezer at -8.5°C and placed in an 85-g aluminum calorimeter filled with 310 g of water at room temperature of 20.0°C. The final situation is all water at 17.0°C. What was the mass of the ice cube?

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