An ice-cream has a marked value of \(700 \mathrm{kcal}\). How many kilo-watt- hour of energy will it deliver to the body as it is digested $(\mathrm{J}=4.2 \mathrm{~J} / \mathrm{cal})$ (A) \(0.81 \mathrm{kwh}\) (B) \(0.90 \mathrm{kwh}\) (C) \(1.11 \mathrm{kwh}\) (D) \(0.71 \mathrm{kwh}\)

Short Answer

Expert verified
The ice-cream delivers 0.82 kWh to the body when digested, and the closest option is (A) 0.81 kWh.

Step by step solution

01

Convert kilocalories to calories

To convert the energy in kilocalories to calories, multiply the given value by 1000: Energy = 700 kcal × 1000 cal/kcal = 700,000 cal
02

Convert calories to Joules

Next, we need to convert this energy from calories to Joules using the provided conversion factor 1 cal = 4.2 J: Energy = 700,000 cal × 4.2 J/cal = 2,940,000 J
03

Convert Joules to kilowatt-hours

Finally, we need to convert the energy from Joules to kilowatt-hours. We know that 1 kWh = 3,600,000 J. Therefore, Energy = \( \frac{2,940,000 \mathrm{J}}{3,600,000 \mathrm{J/kWh}} = 0.8167 \mathrm{kWh} \)
04

Round the answer and choose the correct option

Round the result to two decimal places: Energy = 0.82 kWh Comparing our result to the available options, the closest value is (A) 0.81 kWh.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A bomb of mass \(3.0 \mathrm{~kg}\) explodes in air into two pieces of masses \(2.0 \mathrm{~kg}\) and \(1.0 \mathrm{~kg}\). The smaller mass goes at a speed of \(80 \mathrm{~m} / \mathrm{s}\). The total energy imparted to the two fragments is (A) \(1.07 \mathrm{KJ}\) (B) \(2.14 \mathrm{KJ}\) (C) \(2.4 \mathrm{KJ}\) (D) \(4.8 \mathrm{KJ}\)

An engine pump is used to pump a liquid of density \(\rho\) continuously through a pipe of cross-sectional area \(\mathrm{A}\). If the speed of flow of the liquid in the pipe is \(\mathrm{v}\), then the rate at which kinetic energy is being imparted to the liquid is (A) \((1 / 2) \mathrm{A} \rho \mathrm{V}^{3}\) (B) \((1 / 2) \mathrm{A} \rho \mathrm{V}^{2}\) (C) \((1 / 2) \mathrm{A} \rho \mathrm{V}\) (B) \(\mathrm{ApV}\)

A body of mass \(1 \mathrm{~kg}\) is thrown upwards with a velocity $20 \mathrm{~m} / \mathrm{s}\(. It momentarily comes to rest after a height \)18 \mathrm{~m}\(. How much energy is lost due to air friction. \)(\mathrm{g}=10 \mathrm{~m} / \mathrm{s} 2)$ (A) \(20 \mathrm{~J}\) (B) \(30 \mathrm{~J}\) (C) \(40 \mathrm{~J}\) (D) \(10 \mathrm{~J}\)

A particle is moving under the influence of a force given by \(\mathrm{F}=\mathrm{kx}\), where \(\mathrm{k}\) is a constant and \(\mathrm{x}\) is the distance moved. What energy (in joule) gained by the particle in moving from \(\mathrm{x}=1 \mathrm{~m}\) to \(\mathrm{x}=3 \mathrm{~m} ?\) (A) \(2 \mathrm{k}\) (B) \(3 \mathrm{k}\) (C) \(4 \mathrm{k}\) (D) \(9 \mathrm{k}\)

Assertion and Reason are given in following questions. Each question have four option. One of them is correct it. (1) If both assertion and reason and the reason is the correct explanation of the Assertion. (2) If both assertion and reason are true but reason is not the correct explanation of the assertion. (3) If the assertion is true but reason is false. (4) If the assertion and reason both are false. Assertion: A weight lifter does no work in holding the weight up. Reason: Work done is zero because distance moved is zero. (A) 1 (B) 2 (C) 3 (D) 4

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free