/*! 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 21 A \(2.0 \mathrm{~kg}\) metal blo... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(2.0 \mathrm{~kg}\) metal block \(\left(c=0.137 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) is heated from \(15^{\circ} \mathrm{C}\) to \(90^{\circ} \mathrm{C}\). By how much does its internal energy change?

Short Answer

Expert verified
The internal energy change is 20550 calories.

Step by step solution

01

Identify the Known Values

First, let's identify the known values from the problem. - Mass of the metal block, \( m = 2.0\, \text{kg} \)- Initial temperature, \( T_1 = 15^{\circ} \text{C} \)- Final temperature, \( T_2 = 90^{\circ} \text{C} \)- Specific heat capacity, \( c = 0.137\, \text{cal/g} \cdot { }^{\circ}\text{C} \)Convert the mass to grams:\[ m = 2.0\, \text{kg} = 2000\, \text{g} \]
02

Calculate the Temperature Change

Calculate the change in temperature using:\[ \Delta T = T_2 - T_1 \]Substituting the values:\[ \Delta T = 90^{\circ} \text{C} - 15^{\circ} \text{C} = 75^{\circ} \text{C} \]
03

Calculate the Change in Internal Energy

The change in internal energy (\( \Delta U \)) of the block can be calculated using the formula:\[ \Delta U = m \cdot c \cdot \Delta T \]Substituting the known values:\[ \Delta U = 2000\, \text{g} \times 0.137\, \text{cal/g} \cdot { }^{\circ}\text{C} \times 75^{\circ}\text{C} \]\[ \Delta U = 20550\, \text{cal} \]Thus, the internal energy change is 20550 calories.

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Key Concepts

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

Internal Energy Change
The concept of internal energy change is an important one in thermodynamics. It connects the energetic shifts within a system to temperature and amount of matter. Internal energy, represented often as \( \Delta U \), can be thought of as the sum of all molecular energies within a substance as they move and interact.

When a metal block, for example, is heated, its molecules gain more energy. This increased energy manifests as increased internal energy. To calculate this change, we use the equation:
  • \( \Delta U = m \cdot c \cdot \Delta T \)
Here,
  • \( m \) is the mass of the block,
  • \( c \) is the specific heat capacity,
  • \( \Delta T \) is the change in temperature.
By multiplying these three variables, we can see exactly how much energy was added or lost. Hence, if a solid metal block experiences heating, its internal energy increases because it absorbs heat to fuel the molecular motion.
Temperature Change
Temperature change is a straightforward yet vital concept when dealing with heat and energy problems. It refers to how much warmer or cooler an object becomes. In formulas, this change is denoted by the Greek letter \( \Delta T \).

Calculating \( \Delta T \) is simple:
  • You subtract the initial temperature \( T_1 \) from the final temperature \( T_2 \).
For instance, if a block's temperature goes from \( 15^{\circ} \text{C} \) to \( 90^{\circ} \text{C} \), the temperature change is:
  • \( \Delta T = 90^{\circ} \text{C} - 15^{\circ} \text{C} = 75^{\circ} \text{C} \)
Temperature change is crucial in determining how much energy a material requires to heat or cool, according to its specific heat capacity. It helps us understand how much heat energy needs to be supplied or removed in heating systems or coolants.
Mass Conversion
Mass conversion is an essential step in many physics problems, particularly when units must align correctly for calculations. When solving heat transfer problems like changes in internal energy, converting mass from one unit to another ensures consistency.

In the given problem, the mass of the metal block is originally provided in kilograms (\( \text{kg} \)). Since the specific heat capacity is given in calories per gram \( (\text{cal/g} \cdot { }^{\circ}\text{C}) \), it's necessary to convert the mass to grams as well.
  • The conversion factor is \( 1 \text{ kg} = 1000 \text{ g} \).
  • Therefore, \( 2.0 \text{ kg} = 2000 \text{ g} \).
This conversion allows us to seamlessly integrate mass into the formula for calculating internal energy change, ensuring all measurements are compatible and the calculative result is accurate.

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Most popular questions from this chapter

How much external work is done by an ideal gas in expanding from a volume of \(3.0\) liters to a volume of \(30.0\) liters against a constant pressure of \(2.0\) atm?

An ideal gas expands adiabatically to three times its original volume. In doing so, the gas does \(720 \mathrm{~J}\) of work. (a) How much heat flows from the gas? \((b)\) What is the change in internal energy of the gas? \((c)\) Does its temperature rise or fall?

(a) Compute \(c_{v}\) for the monatomic gas argon, given \(c_{p}=0.125 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and \(\gamma=1.67 .(b)\) Compute \(c_{p}\) for the diatomic gas nitric oxide (NO), given \(c_{v}=0.166 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and \(\gamma=1.40\).

As \(3.0\) liters of ideal gas at \(27^{\circ} \mathrm{C}\) is heated, it expands at a constant pressure of \(2.0 \mathrm{~atm}\). How much work is done by the gas as its temperature is changed from \(27^{\circ} \mathrm{C}\) to \(227^{\circ} \mathrm{C}\) ?

Three kilomoles \((6.00 \mathrm{~kg})\) of hydrogen gas at S.T.P. expands isobarically to precisely twice its volume. (a) What is the final temperature of the gas? (b) What is the expansion work done by the gas? (c) By how much does the internal energy of the gas change? \((d)\) How much heat enters the gas during the expansion? For \(\mathrm{H}_{2}, c_{v}=10.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\). Assume the hydrogen will behave as an ideal gas. (a) From \(P_{1} V_{1} / T_{1}=P_{2} V_{2} / T_{2}\) with \(P_{1}=P_{2}\), $$T_{2}=T_{1}\left(\frac{V_{2}}{V_{1}}\right)=(273 \mathrm{~K})(2.00)=546 \mathrm{~K}$$ (b) Because \(1 \mathrm{kmol}\) at S.T.P. occupies \(22.4 \mathrm{~m}^{3}\), we have \(V_{1}=67.2 \mathrm{~m}^{3}\). Then $$\Delta W=P \Delta V=P\left(V_{2}-V_{1}\right)=\left(1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(67.2 \mathrm{~m}^{3}\right)=6.8 \mathrm{MJ}$$ (c) To raise the temperature of this ideal gas by \(273 \mathrm{~K}\) at constant volume requires $$\Delta Q=c_{v} m \Delta T=(10.0 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(6.00 \mathrm{~kg})(273 \mathrm{~K})=16.4 \mathrm{MJ}$$ Because the volume is constant here, no work is done and \(\Delta Q\) equals the internal energy that must be added to the \(6.00 \mathrm{~kg}\) of \(\mathrm{H}_{2}\) to change its temperature from \(273 \mathrm{~K}\) to \(546 \mathrm{~K}\). Therefore, \(\Delta U=16.4 \mathrm{MJ}\). (d) The system obeys the First Law during the process and so $$ \Delta Q=\Delta U+\Delta W=16.4 \mathrm{MJ}+6.8 \mathrm{MJ}=23.2 \mathrm{MJ} $$

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