Ch 17: Temperature and Heat

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 17: Temperature and Heat

Problem 51

Chapter 17, Problem 51

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.)

Verified step by step guidance

Calculate the kinetic energy (KE) of the asteroid using the formula: $ KE = \frac{1}{2}mv^2 $, where $ m $ is the mass of the asteroid and $ v $ is its velocity. Convert the velocity from km/s to m/s before substituting into the formula.

Determine the amount of kinetic energy that goes into boiling the water. Since 1.00% of the asteroid's kinetic energy is used, multiply the total kinetic energy by 0.01.

Calculate the energy required to boil water using the formula: $ Q = mL + mc\Delta T $, where $ m $ is the mass of water, $ L $ is the latent heat of vaporization, $ c $ is the specific heat capacity of water, and $ \Delta T $ is the change in temperature. Assume $ L = 2.26 \times 10^6 $ J/kg and $ c = 4.18 \times 10^3 $ J/(kg°C).

Set the energy from the asteroid equal to the energy required to boil the water: $ 0.01 \times KE = mL + mc\Delta T $. Solve for $ m $, the mass of water boiled away.

Rearrange the equation to solve for $ m $: $ m = \frac{0.01 \times KE}{L + c\Delta T} $. Substitute the known values to find the mass of water that will be boiled away.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

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

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion, calculated using the formula KE = 0.5 * m * v^2, where m is mass and v is velocity. In this scenario, the asteroid's kinetic energy is crucial for determining how much energy is transferred to the ocean water upon impact.

Heat Transfer and Phase Change

Heat transfer involves the movement of thermal energy from one object to another, often resulting in a change of state, such as boiling. The energy required to boil water is determined by its specific heat and latent heat of vaporization, which are essential for calculating the mass of water boiled away by the asteroid's impact.

Energy Conversion Efficiency

Energy conversion efficiency refers to the percentage of energy that is successfully converted from one form to another. In this problem, only 1.00% of the asteroid's kinetic energy is used to boil the ocean water, highlighting the importance of understanding how energy is distributed and utilized in physical processes.

Recommended video:

Guided course

07:46

Unit Conversions

Related Practice

Textbook Question

A laboratory technician drops a $0.0850$ -kg sample of unknown solid material, at $100.0$ °C, into a calorimeter. The calorimeter can, initially at $19.0$ °C, is made of $0.150$ kg of copper and contains $0.200$ kg of water. The final temperature of the calorimeter can and contents is $26.1$ °C. Compute the specific heat of the sample.

views

Textbook Question

A $4.00$ -kg silver ingot is taken from a furnace, where its temperature is $750.0$ °C, and placed on a large block of ice at $0.0$ °C. Assuming that all the heat given up by the silver is used to melt the ice, how much ice is melted?

views

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

Before going in for his annual physical, a 70.0 kg man whose body temperature is 37.0°C consumes an entire 0.355-L can of a soft drink (mostly water) at 12.0°C. What will his body temperature be after equilibrium is attained? Ignore any heating by the man’s metabolism. The specific heat of the man’s body is 3480 J/kg K.

Textbook Question

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.

views

Textbook Question

An insulated beaker with negligible mass contains $0.250$ kg of water at $75.0$ °C. How many kilograms of ice at $negative 20.0$ °C must be dropped into the water to make the final temperature of the system $40.0$ °C?

views