Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 79

The amount of heat generated during the contraction of muscle in an amphibian's leg is given by $$Q=0.544 \mathrm{mJ}+(1.46 \mathrm{mJ} / \mathrm{cm}) \Delta x$$ where \(\Delta x\) is the length shortened. If a muscle of length \(3.0 \mathrm{cm}\) and mass \(0.10 \mathrm{g}\) is shortened by $1.5 \mathrm{cm}$ during a contraction, what is the temperature rise? Assume that the specific heat of muscle is $4.186 \mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right)$.

Short Answer

Expert verified
Question: Determine the temperature rise during the amphibian muscle's contraction. Answer: The temperature rise during the amphibian muscle's contraction is approximately \(0.00653^{\circ} \mathrm{C}\).

Step by step solution

01

Find the heat generated by muscle contraction

We are given the formula for the amount of heat generated during the contraction: $$Q=0.544 \mathrm{mJ}+(1.46 \mathrm{mJ} / \mathrm{cm}) \Delta x$$ The muscle is shortened by \(\Delta x = 1.5 \mathrm{cm}\). Substituting \(\Delta x\) into the formula, we get: $$Q=0.544 \mathrm{mJ}+(1.46 \mathrm{mJ} / \mathrm{cm}) (1.5 \mathrm{cm})$$
02

Calculate the heat generated

To calculate the heat generated during the muscle contraction, perform the calculations: $$Q=0.544 \mathrm{mJ}+(1.46 \mathrm{mJ} / \mathrm{cm}) (1.5 \mathrm{cm})$$ $$Q=(0.544 + (1.46)(1.5)) \mathrm{mJ}$$ $$Q=(0.544 + 2.19) \mathrm{mJ}$$ $$Q=2.734 \mathrm{mJ}$$ Convert this value to Joules (1 mJ = 0.001 J): $$Q=0.002734 \mathrm{J}$$
03

Use the specific heat to find the temperature rise

We are given the specific heat of muscle, \(C_m = 4.186 \mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right)\), and the mass of the muscle, \(m = 0.10 \mathrm{g}\). The temperature rise, \(\Delta T\), can be found using the formula: $$Q = mC_m\Delta T$$ Rearranging this formula to solve for \(\Delta T\): $$\Delta T=\frac{Q}{mC_m}$$ Now, substitute the values of \(Q, m,\) and \(C_m\) into the formula: $$\Delta T=\frac{0.002734 \mathrm{J}}{0.10 \mathrm{g} \times 4.186 \mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right)}$$
04

Calculate the temperature rise

Finish the calculation to determine the temperature rise during muscle contraction: $$\Delta T=\frac{0.002734 \mathrm{J}}{0.10 \mathrm{g} \times 4.186 \mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right)}$$ $$\Delta T=\frac{0.002734}{0.4186}^{\circ} \mathrm{C}$$ $$\Delta T \approx 0.00653^{\circ} \mathrm{C}$$
05

State the final answer

The temperature rise during the amphibian muscle's contraction is approximately \(0.00653^{\circ} \mathrm{C}\).

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 75 -kg block of ice at \(0.0^{\circ} \mathrm{C}\) breaks off from a glacier, slides along the frictionless ice to the ground from a height of $2.43 \mathrm{m},$ and then slides along a horizontal surface consisting of gravel and dirt. Find how much of the mass of the ice is melted by the friction with the rough surface, assuming \(75 \%\) of the internal energy generated is used to heat the ice.
It is a damp, chilly day in a New England seacoast town suffering from a power failure. To warm up the cold, clammy sheets, Jen decides to fill hot water bottles to tuck between the sheets at the foot of the beds. If she wishes to heat \(2.0 \mathrm{L}\) of water on the wood stove from $20.0^{\circ} \mathrm{C}\( to \)80.0^{\circ} \mathrm{C},$ how much heat must flow into the water?
A mass of \(1.00 \mathrm{kg}\) of water at temperature \(T\) is poured from a height of \(0.100 \mathrm{km}\) into a vessel containing water of the same temperature \(T,\) and a temperature change of \(0.100^{\circ} \mathrm{C}\) is measured. What mass of water was in the vessel? Ignore heat flow into the vessel, the thermometer, etc.
A thermometer containing \(0.10 \mathrm{g}\) of mercury is cooled from \(15.0^{\circ} \mathrm{C}\) to \(8.5^{\circ} \mathrm{C} .\) How much energy left the mercury in this process?
A person of surface area \(1.80 \mathrm{m}^{2}\) is lying out in the sunlight to get a tan. If the intensity of the incident sunlight is $7.00 \times 10^{2} \mathrm{W} / \mathrm{m}^{2},$ at what rate must heat be lost by the person in order to maintain a constant body temperature? (Assume the effective area of skin exposed to the Sun is \(42 \%\) of the total surface area, \(57 \%\) of the incident radiation is absorbed, and that internal metabolic processes contribute another \(90 \mathrm{W}\) for an inactive person.)
See all solutions

Recommended explanations on Physics Textbooks

Physical Quantities And Units

Read Explanation

Nuclear Physics

Read Explanation

Thermodynamics

Read Explanation

Particle Model of Matter

Read Explanation

Work Energy and Power

Read Explanation

Scientific Method 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