Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 4: Problem 170

Consider a 7.6-cm-long and 3-cm-diameter cylindrical lamb meat chunk \(\left(\rho=1030 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=3.49 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\right.\), \(\left.k=0.456 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=1.3 \times 10^{-7} \mathrm{~m}^{2} / \mathrm{s}\right)\). Fifteen such meat chunks initially at \(2^{\circ} \mathrm{C}\) are dropped into boiling water at \(95^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(1200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The amount of heat transfer during the first 8 minutes of cooking is (a) \(71 \mathrm{~kJ}\) (b) \(227 \mathrm{~kJ}\) (c) \(238 \mathrm{~kJ}\) \(\begin{array}{ll}\text { (d) } 269 \mathrm{~kJ} & \text { (e) } 307 \mathrm{~kJ}\end{array}\)

Short Answer

Expert verified
Answer: (c) 238 kJ

Step by step solution

01

Understand the formula for heat transfer

The formula for the heat transfer (Q) between the meatchunk and boiling water is given by the equation: Q = h * A * ΔT * t where: h = heat transfer coefficient A = surface area ΔT = Temperature difference t = time. We will calculate the parameters based on the given information.
02

Find the surface area of the meat chunk

The meat is cylindrical in shape. The formula for surface area (A) of a cylinder can be written as: A = 2 * π * r * (r + l) Given: length(l) = 7.6 cm, diameter = 3 cm which means radius(r) = 1.5 cm. Let's find the surface area (A): A = 2 * π * 1.5 * (1.5 + 7.6) = 2 * π * 1.5 * 9.1 = 27.3π cm² Since we want the surface area in meters, we convert the cm² to m²: A = 27.3π × (0.01)² = 0.0273π m²
03

Find the temperature difference ΔT between the meat chunks and boiling water

Given: initial temperature of meat = 2°C, temperature of the boiling water = 95°C Temperature Difference (ΔT) = Temperature of the boiling water - initial temperature of meat ΔT = 95 - 2 = 93°C
04

Find the time in seconds

Given: time t = 8 minutes. We need to convert the time to seconds. t = 8 × 60 = 480 seconds
05

Calculate the heat transfer Q for one meat chunk

Now we have all the parameters to calculate the heat transfer (Q) for one meat chunk: h = 1200 W/m²K A = 0.0273π m² ΔT = 93°C t = 480 seconds Q = h * A * ΔT * t Q = 1200 * 0.0273π *93 * 480 Q = 1200 * 0.0273π * 93 *480 ≈ 15.872 kJ.
06

Calculate the heat transfer Q for all 15 meat chunks

Since there are 15 meat chunks dropped into the boiling water, we need to multiply the heat transfer for one chunk by 15 to get the total heat transfer: Q(total) = Q * 15 Q(total) = 15.872 * 15 ≈ 238.08 kJ Now we will compare the obtained value with the given options. a) 71 kJ b) 227 kJ c) 238 kJ d) 269 kJ e) 307 kJ The closest option to our calculated value of heat transfer is (c) 238 kJ. So, the heat transfer during the first 8 minutes of cooking is approximately 238 kJ.

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.

Start your free trial

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

Oxy-fuel combustion power plants use pulverized coal particles as fuel to burn in a pure oxygen environment to generate electricity. Before entering the furnace, pulverized spherical coal particles with an average diameter of \(300 \mu \mathrm{m}\), are being transported at \(2 \mathrm{~m} / \mathrm{s}\) through a \(3-\mathrm{m}\) long heated tube while suspended in hot air. The air temperature in the tube is \(900^{\circ} \mathrm{C}\) and the average convection heat transfer coefficient is \(250 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperature of the coal particles at the exit of the heated tube, if the initial temperature of the particles is \(20^{\circ} \mathrm{C}\).

Consider a curing kiln whose walls are made of \(30-\mathrm{cm}-\) thick concrete with a thermal diffusivity of \(\alpha=0.23 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\). Initially, the kiln and its walls are in equilibrium with the surroundings at \(6^{\circ} \mathrm{C}\). Then all the doors are closed and the kiln is heated by steam so that the temperature of the inner surface of the walls is raised to \(42^{\circ} \mathrm{C}\) and the temperature is maintained at that level for \(2.5 \mathrm{~h}\). The curing kiln is then opened and exposed to the atmospheric air after the steam flow is turned off. If the outer surfaces of the walls of the kiln were insulated, would it save any energy that day during the period the kiln was used for curing for \(2.5 \mathrm{~h}\) only, or would it make no difference? Base your answer on calculations.

Lumped system analysis of transient heat conduction situations is valid when the Biot number is (a) very small (b) approximately one (c) very large (d) any real number (e) cannot say unless the Fourier number is also known.

4-115 A semi-infinite aluminum cylinder \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), \(\left.\alpha=9.71 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right)\) of diameter \(D=15 \mathrm{~cm}\) is initially at a uniform temperature of \(T_{i}=115^{\circ} \mathrm{C}\). The cylinder is now placed in water at \(10^{\circ} \mathrm{C}\), where heat transfer takes place by convection with a heat transfer coefficient of \(h=140 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the temperature at the center of the cylinder \(5 \mathrm{~cm}\) from the end surface 8 min after the start of cooling. 4-116 A 20-cm-long cylindrical aluminum block \((\rho=\) \(2702 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=0.896 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}, k=236 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\alpha=\) \(\left.9.75 \times 10^{-5} \mathrm{~m}^{2} / \mathrm{s}\right), 15 \mathrm{~cm}\) in diameter, is initially at a uniform temperature of \(20^{\circ} \mathrm{C}\). The block is to be heated in a furnace at \(1200^{\circ} \mathrm{C}\) until its center temperature rises to \(300^{\circ} \mathrm{C}\). If the heat transfer coefficient on all surfaces of the block is \(80 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine how long the block should be kept in the furnace. Also, determine the amount of heat transfer from the aluminum block if it is allowed to cool in the room until its temperature drops to \(20^{\circ} \mathrm{C}\) throughout.

Consider the freezing of packaged meat in boxes with refrigerated air. How do \((a)\) the temperature of air, (b) the velocity of air, \((c)\) the capacity of the refrigeration system, and \((d)\) the size of the meat boxes affect the freezing time?
See all solutions

Recommended explanations on Physics Textbooks

Famous Physicists

Read Explanation

Space Physics

Read Explanation

Astrophysics

Read Explanation

Fundamentals of Physics

Read Explanation

Fields in Physics

Read Explanation

Rotational Dynamics

Read Explanation
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