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Ch 17: Temperature and Heat
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 17, Problem 42

An ice-cube tray of negligible mass contains 0.290 kg of water at 18.0°C. How much heat must be removed to cool the water to 0.00°C and freeze it? Express your answer in joules, calories, and Btu.

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First, calculate the heat required to cool the water from 18.0°C to 0.0°C using the formula for sensible heat: Q=mcΔT, where m is the mass of the water, c is the specific heat capacity of water (4.186 J/g°C), and ΔT is the change in temperature.
Convert the mass of water from kilograms to grams, since the specific heat capacity is given in J/g°C. Use the conversion: m=0.290 kg × 1000 g/kg.
Calculate the heat required to freeze the water at 0.0°C using the formula for latent heat: Q=mL, where L is the latent heat of fusion for water (334 J/g).
Add the heat calculated for cooling and freezing to find the total heat removed. This total heat will be the sum of the sensible heat and the latent heat.
Convert the total heat removed from joules to calories and Btu using the conversion factors: 1 calorie = 4.184 joules and 1 Btu = 1055 joules.

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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 change the temperature of a unit mass of a substance by one degree Celsius. For water, this value is 4.18 J/g°C. It is crucial for calculating the heat needed to lower the temperature of the water from 18.0°C to 0.0°C before freezing.
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Latent Heat of Fusion

Latent heat of fusion is the heat required to change a unit mass of a substance from solid to liquid (or vice versa) at constant temperature. For water, this is 334 J/g. This concept is essential for determining the heat removal needed to freeze the water once it reaches 0.0°C.
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Heat Transfer and Units

Heat transfer involves the movement of thermal energy from one object to another. It is measured in joules (J), calories (cal), and British thermal units (Btu). Understanding these units is necessary to express the total heat removed in different measurement systems, as required by the question.
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Related Practice
Textbook Question

A copper calorimeter can with mass 0.100 kg contains 0.160 kg of water and 0.0180 kg of ice in thermal equilibrium at atmospheric pressure. If 0.750 kg of lead at 255°C is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings.

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Textbook Question

A laboratory technician drops a 0.08500.0850-kg sample of unknown solid material, at 100.0100.0°C, into a calorimeter. The calorimeter can, initially at 19.019.0°C, is made of 0.1500.150 kg of copper and contains 0.2000.200 kg of water. The final temperature of the calorimeter can and contents is 26.126.1°C. Compute the specific heat of the sample.

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Textbook Question

An asteroid with a diameter of 10 km and a mass of 2.60 × 1015 kg impacts the earth at a speed of 32.0 km/s, landing in the Pacific Ocean. If 1.00% of the asteroid's kinetic energy goes to boiling the ocean water (assume an initial water temperature of 10.0°C), what mass of water will be boiled away by the collision? (For comparison, the mass of water contained in Lake Superior is about 2 × 1015 kg.)

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Textbook Question

A copper pot with a mass of 0.500 kg contains 0.170 kg of water, and both are at 20.0°C. A 0.250-kg block of iron at 85.0°C is dropped into the pot. Find the final temperature of the system, assuming no heat loss to the surroundings.

Textbook Question

A blacksmith cools a 1.20 kg chunk of iron, initially at 650.0°C, by trickling 15.0°C water over it. All of the water boils away, and the iron ends up at 120.0°C. How much water did the blacksmith trickle over the iron?

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Textbook Question

An insulated beaker with negligible mass contains 0.2500.250 kg of water at 75.075.0°C. How many kilograms of ice at 20.0-20.0°C must be dropped into the water to make the final temperature of the system 40.040.0°C?

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