Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 8

A material has mass density \(\rho,\) volume \(V\), and specific heat \(c .\) Which of the following is a correct expression for the heat exchange that occurs when the material's temperature changes by \(\Delta T\) in degrees Celsius? a) \((\rho c / V) \Delta T\) b) \((\rho c V)(\Delta T+273.15)\) c) \((\rho c V) / \Delta T\) d) \(\rho c V \Delta T\)

Short Answer

Expert verified
a) (ρc / V) ΔT b) (ρcV)(ΔT+273.15) c) (ρcV) / ΔT d) ρ c V ΔT Answer: d) ρ c V ΔT

Step by step solution

01

Write down the equation for heat exchange

The formula for calculating the heat exchange, \(Q\), when a material's temperature changes by \(\Delta T\) is: \(Q = mc\Delta T\)
02

Replace mass with mass density and volume

Since mass \(m\) can be written as the product of mass density (\(\rho\)) and volume (\(V\)): \(m = \rho V\) Now substitute this into the heat exchange equation from Step 1: \(Q = (\rho V)c\Delta T\) Now, we need to find the option that matches the above expression.
03

Compare with the given options

Let's look at each option: a) \((\rho c / V) \Delta T\) - This option divides \(\rho c\) by \(V\), which is not the same as our obtained expression. b) \((\rho c V)(\Delta T+273.15)\) - This option adds 273.15 to the temperature change \(\Delta T\). This is not in our obtained expression. c) \((\rho c V) / \Delta T\) - This divides \(\rho c V\) by \(\Delta T\), while our expression has \(\rho V c \Delta T\). d) \(\rho c V \Delta T\) - This option matches our obtained expression, which is the correct formula for the heat exchange when the temperature changes by \(\Delta T\). So the correct answer is: d) \(\rho c V \Delta T\)

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

A metal brick found in an excavation was sent to a testing lab for nondestructive identification. The lab weighed the sample brick and found its mass to be \(3.0 \mathrm{~kg} .\) The brick was heated to a temperature of \(3.0 \cdot 10^{2}{ }^{\circ} \mathrm{C}\) and dropped into an insulated copper calorimeter of mass 1.5 kg containing \(2.0 \mathrm{~kg}\) of water at \(2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C} .\) The final temperature at equilibrium was noted to be \(31.7^{\circ} \mathrm{C}\). By calculating the specific heat of the sample from this data, can you identify the brick's material?

You were lost while hiking outside wearing only a bathing suit. a) Calculate the power radiated from your body, assuming that your body's surface area is about \(2.00 \mathrm{~m}^{2}\) and your skin temperature is about \(33.0^{\circ} \mathrm{C} .\) Also, assume that your body has an emissivity of 1.00 . b) Calculate the net radiated power from your body when you were inside a shelter at \(20.0^{\circ} \mathrm{C}\). c) Calculate the net radiated power from your body when your skin temperature dropped to \(27.0^{\circ} \mathrm{C}\).

Determine the ratio of the heat flow into a six-pack of aluminum soda cans to the heat flow into a 2.00 - \(\mathrm{L}\) plastic bottle of soda when both are taken out of the same refrigerator, that is, have the same initial temperature difference with the air in the room. Assume that each soda can has a diameter of \(6.00 \mathrm{~cm}\), a height of \(12.0 \mathrm{~cm}\), and a thickness of \(0.100 \mathrm{~cm}\). Use \(205 \mathrm{~W} /(\mathrm{m} \mathrm{K})\) as the thermal conductivity of aluminum. Assume that the 2.00 - \(\mathrm{L}\) bottle of soda has a diameter of \(10.0 \mathrm{~cm}\), a height of \(25.0 \mathrm{~cm}\), and a thickness of \(0.100 \mathrm{~cm} .\) Use \(0.100 \mathrm{~W} /(\mathrm{mK})\) as the thermal conductivity of plastic.

The label on a soft drink states that 12 fl. oz \((355 \mathrm{~g})\) provides \(150 \mathrm{kcal}\). The drink is cooled to \(10.0^{\circ} \mathrm{C}\) before it is consumed. It then reaches body temperature of \(37^{\circ} \mathrm{C} .\) Find the net energy content of the drink. (Hint: You can treat the soft drink as being identical to water in terms of heat capacity.)

A 1.19-kg aluminum pot contains 2.31 L of water. Both pot and water are initially at \(19.7^{\circ} \mathrm{C} .\) How much heat must flow into the pot and the water to bring their temperature up to \(95.0^{\circ} \mathrm{C}\) ? Assume that the effect of water evaporation during the heating process can be neglected and that the temperature remains uniform throughout the pot and the water.
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Engineering Physics

Read Explanation

Measurements

Read Explanation

Electromagnetism

Read Explanation

Geometrical and Physical Optics

Read Explanation

Waves Physics

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