/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 A \(5.0-\mathrm{kg}\) block of a... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 25

A \(5.0-\mathrm{kg}\) block of aluminum is heated from \(20^{\circ} \mathrm{C}\) to \(90^{\circ} \mathrm{C}\) at atmospheric pressure. Find (a) the work done by the aluminum, (b) the amount of energy transferred to it by heat, and (c) the increase in its internal energy.

Short Answer

Expert verified
The work done by the aluminum is 0J, the energy transferred to it by heat is 315,000J or 315kJ, and the increase in its internal energy is also 315,000J or 315kJ.

Step by step solution

01

Calculate the Change in Temperature

Firstly, let's calculate the change in temperature which is given by \( \Delta T = T_{final} - T_{initial} \) where \( T_{final} = 90^{\circ}\mathrm{C} \) and \( T_{initial} = 20^{\circ}\mathrm{C} \). So, \( \Delta T = 90^{\circ}\mathrm{C} - 20^{\circ}\mathrm{C} = 70^{\circ}\mathrm{C} = 70 K \) (since a difference of degrees Celsius is equivalent to the difference in Kelvin).
02

Calculate the Work Done by the Aluminum Block

The work done by the system is equal to zero because the aluminum block is a solid and does not change volume under atmospheric pressure. Hence, Work done (W) = 0 J.
03

Calculate Energy Transferred as Heat

The energy transferred to the aluminum block as heat can be calculated by the formula \( Q = mc\Delta T \) where \( m \) is the mass of the aluminum block (5.0kg), \( c \) is the specific heat capacity of Aluminum (\(900 \, J/(kg \cdot K)\)) and \( \Delta T \) is the change in temperature (70K). Hence, \( Q = 5.0kg \cdot 900 \, J/(kg \cdot K) \cdot 70K = 315,000J \) or \( 315kJ \).
04

Calculate the Increase in Internal Energy

The increase in the internal energy can be obtained by using the first law of thermodynamics which states that \( \Delta U = Q - W \), where \( \Delta U \) is the change in internal energy, \( Q \) is the heat transferred to the system, and \( W \) is the work done by the system. Substituting the given values, we get \( \Delta U = 315,000J - 0J = 315,000J \) or \( 315kJ \).

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Most popular questions from this chapter

\(\mathrm{M}\) One mole of neon gas is heated from \(300 \mathrm{~K}\) to \(420 \mathrm{~K}\) at constant pressure. Calculate (a) the energy \(Q\) transferred to the gas, (b) the change in the internal energy of the gas, and (c) the work done on the gas. Note that neon has a molar specific heat of \(c=20.79 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\) for a constant-pressure process.

A heat engine operates between a reservoir at \(25^{\circ} \mathrm{C}\) and one at \(375^{\circ} \mathrm{C}\). What is the maximum efficiency possible for this engine?

A \(100-\mathrm{kg}\) steel support rod in a building has a length of \(2.0 \mathrm{~m}\) at a temperature of \(20^{\circ} \mathrm{C}\). The rod supports a hanging load of \(6000 \mathrm{~kg}\). Find (a) the work done on the rod as the temperature increases to \(40^{\circ} \mathrm{C}\), (b) the energy \(Q\) added to the rod (assume the specific heat of steel is the same as that for iron), and (c) the change in internal energy of the rod.

\(\mathrm{M}\) A heat engine operates in a Carnot cycle between \(80.0^{\circ} \mathrm{C}\) and \(350^{\circ} \mathrm{C}\). It absorbs \(21000 \mathrm{~J}\) of energy per cycle from the hot reservoir. The duration of each cycle is \(1.00 \mathrm{~s}\). (a) What is the mechanical power output of this engine? (b) How much energy does it expel in cach cycle by heat? 12.5 Entropy

In one cycle a heat engine absorbs \(500 \mathrm{~J}\) from a hightemperature reservoir and expels \(300 \mathrm{~J}\) to a low temperature reservoir. If the efficiency of this engine is \(60 \%\) of the efficiency of a Carnot engine, what is the ratio of the low temperature to the high temperature in the Carnot engine?
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