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Chapter 20: Problem 3

The temperature of a silver bar rises by \(10.0^{\circ} \mathrm{C}\) when it absorbs \(1.23 \mathrm{kJ}\) of energy by heat. The mass of the bar is \(525 \mathrm{g} .\) Determine the specific heat of silver.

Short Answer

Expert verified
The specific heat of silver is approximately \(0.235 \, J/g \cdot \degree{C}\) or \(235 \, J/kg \cdot \degree{C}\).

Step by step solution

01

Convert the Energy

First, convert the kjoules of energy absorbed to joules: \(1.23 \, kJ = 1.23 \times 10^{3} \, Joules\) Since 1 kJ = \(10^{3} \, J\).
02

Convert the Mass

Convert the grams of the silver bar to kilograms: \(525 \, g = 0.525 \, kg\). Since 1 kg = \(10^{3} \, g\).
03

Calculate Specific Heat

Substitute the given values and solve for the specific heat \(c\). Rearrange the formula to find \(c\): \(c = \frac{q}{{m \cdot \Delta T}}\). Substituting the values gives: \(c = \frac{1.23 \times 10^{3} \, J}{{0.525 \, kg \times 10.0^{\circ} C}}\). Calculate the result to find the specific heat.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Conversion
When dealing with heat and energy calculations, understanding energy conversion is crucial. In physics, energy can be measured in various units.
In this problem, energy is given in kilojoules (kJ), but we often need to convert it into joules (J) to use in calculations with the specific heat formula.
The conversion is simple: remember that 1 kilojoule is equal to 1,000 joules. Hence, to convert kilojoules to joules, you multiply the energy in kilojoules by 1,000.
For example, in this problem, energy absorbed by the silver bar is 1.23 kJ, which converts to:- \(1.23 \, \text{kJ} = 1.23 \times 10^3 \, \text{J}\)Through this conversion, the energy is now compatible with the joule-based formula for specific heat calculations.
Temperature Change
Temperature change is a significant factor when calculating specific heat. It reflects how much the temperature of a substance alters when energy is applied.
In the specific heat formula, the temperature change is denoted as \(\Delta T\), which is simply the difference between the final and initial temperatures of the substance.
In this problem, the silver bar experiences a temperature rise by \(10.0^{\circ}\text{C}\).
This direct change indicates how much energy is needed to increase the temperature of a particular mass.- The role of temperature change is a key determinant in understanding how different substances react to heat.- For identical energy input, substances with smaller temperature changes have higher specific heats, meaning they store more energy.
Silver Properties
Properties of materials, such as silver, play a vital role in heat calculations. Different materials absorb and conduct heat differently.
Silver is known for its excellent conductivity and relatively low specific heat compared to other metals. Specific heat is defined as the amount of energy required to raise the temperature of one kilogram of a substance by one degree Celsius.- In the specific heat equation, \(c = \frac{q}{m \cdot \Delta T}\), where \(c\) is specific heat, \(q\) is the heat absorbed, and \(m\) is the mass.- By knowing the mass, energy absorbed, and temperature change, we can calculate the specific heat.- Silver's low specific heat reveals it heats up quickly compared to materials with higher specific heats.This property makes silver efficient in applications where rapid heat transfer is essential, such as in electronics and thermal conductors.

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Most popular questions from this chapter

A flow calorimeter is an apparatus used to measure the specific heat of a liquid. The technique of flow calorimetry involves measuring the temperature difference between the input and output points of a flowing stream of the liquid while energy is added by heat at a known rate. A liquid of density \(\rho\) flows through the calorimeter with volume flow rate \(R\). At steady state, a temperature difference \(\Delta T\) is established between the input and output points when energy is supplied at the rate \(\mathscr{P}\). What is the specific heat of the liquid?

(a) In air at \(0^{\circ} \mathrm{C},\) a \(1.60-\mathrm{kg}\) copper block at \(0^{\circ} \mathrm{C}\) is set sliding at \(2.50 \mathrm{m} / \mathrm{s}\) over a sheet of ice at \(0^{\circ} \mathrm{C} .\) Friction brings the block to rest. Find the mass of the ice that melts. To describe the process of slowing down, identify the energy input \(Q,\) the work input \(W,\) the change in internal energy \(\Delta E_{\text {int }},\) and the change in mechanical energy \(\Delta K\) for the block and also for the ice. (b) A \(1.60-\mathrm{kg}\) block of ice at \(0^{\circ} \mathrm{C}\) is set sliding at \(2.50 \mathrm{m} / \mathrm{s}\) over a sheet of copper at \(0^{\circ} \mathrm{C}\) Friction brings the block to rest. Find the mass of the ice that melts. Identify \(Q, W, \Delta E_{\text {int }},\) and \(\Delta K\) for the block and for the metal sheet during the process. (c) A thin 1.60-kg slab of copper at \(20^{\circ} \mathrm{C}\) is set sliding at \(2.50 \mathrm{m} / \mathrm{s}\) over an identical stationary slab at the same temperature. Friction quickly stops the motion. If no energy is lost to the environment by heat, find the change in temperature of both objects. Identify \(Q, W, \Delta E_{\text {int }},\) and \(\Delta K\) for each object during the process.

During periods of high activity, the Sun has more sunspots than usual. Sunspots are cooler than the rest of the luminous layer of the Sun's atmosphere (the photosphere). Paradoxically, the total power output of the active Sun is not lower than average but is the same or slightly higher than average. Work out the details of the following crude model of this phenomenon. Consider a patch of the photosphere with an area of \(5.10 \times 10^{14} \mathrm{m}^{2} .\) Its emissivity is 0.965 (a) Find the power it radiates if its temperature is uniformly \(5800 \mathrm{K},\) corresponding to the quiet Sun. (b) To represent a sunspot, assume that \(10.0 \%\) of the area is at \(4800 \mathrm{K}\) and the other \(90.0 \%\) is at \(5890 \mathrm{K}\). That is, a section with the surface area of the Earth is \(1000 \mathrm{K}\) cooler than before and a section nine times as large is 90 K warmer. Find the average temperature of the patch. (c) Find the power output of the patch. Compare it with the answer to part (a). (The next sunspot maximum is expected around the year \(2012 .\) )

Two speeding lead bullets, each of mass \(5.00 \mathrm{g},\) and at temperature \(20.0^{\circ} \mathrm{C},\) collide head-on at speeds of \(500 \mathrm{m} / \mathrm{s}\) each. Assuming a perfectly inelastic collision and no loss of energy by heat to the atmosphere, describe the final state of the two-bullet system.

Water in an electric teakettle is boiling. The power absorbed by the water is \(1.00 \mathrm{kW}\). Assuming that the pressure of vapor in the kettle equals atmospheric pressure, determine the speed of effusion of vapor from the kettle's spout, if the spout has a cross-sectional area of \(2.00 \mathrm{cm}^{2}.\)
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