/*! 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 2 The specific heat of water is \(... [FREE SOLUTION] | 91Ó°ÊÓ

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The specific heat of water is \(4184 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). By how many joules does the internal energy of \(50 \mathrm{~g}\) of water change as it is heated from \(21^{\circ} \mathrm{C}\) to \(37^{\circ} \mathrm{C}\) ? Assume that the expansion of the water is negligible. The heat added to raise the temperature of the water is \(\Delta Q=c m \Delta T=(4184 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K})(0.050 \mathrm{~kg})\left(16^{\circ} \mathrm{C}\right)=3.4 \times 10^{3} \mathrm{~J}\) Notice that \(\Delta T\) in Celsius is equal to \(\Delta T\) in kelvin. If we ignore the slight expansion of the water, no work was done on the surroundings and so \(\Delta W=0\). Then, the First Law, \(\Delta Q=\Delta U+\Delta W\), tells us that $$ \Delta U=\Delta Q=3.4 \mathrm{~kJ} $$

Short Answer

Expert verified
The internal energy change is \(3.4 \times 10^3 \, \text{J}\).

Step by step solution

01

Understanding the Problem

We need to calculate the change in internal energy of water when it is heated. We are given the specific heat capacity of water as \(4184 \, \text{J/kg} \cdot \text{K}\) and the mass and temperature change of the water. The internal energy change is calculated using the formula \(\Delta U = \Delta Q\), where \(\Delta Q = cm\Delta T\).
02

Calculate the Mass in Kilograms

The mass of the water is provided as \(50 \, \text{g}\). We need to convert this mass to kilograms for our calculation. To convert grams to kilograms, divide by 1000. Thus, the mass \(m = \frac{50}{1000} = 0.050 \, \text{kg}\).
03

Determine the Temperature Change

The water is heated from \(21^{\circ} \text{C}\) to \(37^{\circ} \text{C}\). The temperature change \(\Delta T\) is \(37 - 21 = 16^{\circ} \text{C}\). Since \(\Delta T\) in Celsius is equivalent to \(\Delta T\) in Kelvin, we have \(\Delta T = 16 \text{ K}\).
04

Calculate Heat Added Using the Specific Heat Formula

The formula to calculate heat added is \(\Delta Q = c m \Delta T\). Substituting in the given values, we have:\[\Delta Q = (4184 \, \text{J/kg} \cdot \text{K})(0.050 \, \text{kg})(16 \, \text{K})\].Carry out the multiplication: \(4184 \times 0.050 \times 16 = 3347.2 \, \text{J}\). Since we are rounding to two significant figures, \(\Delta Q \approx 3.4 \times 10^3 \, \text{J}\).
05

Apply the First Law of Thermodynamics

According to the First Law of Thermodynamics, \(\Delta Q = \Delta U + \Delta W\). Given that the expansion of water is negligible, \(\Delta W = 0\), so \(\Delta U = \Delta Q\). Thus, the change in internal energy \(\Delta U = 3.4 \times 10^3 \, \text{J}\).

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Key Concepts

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

First Law of Thermodynamics
The First Law of Thermodynamics is a fundamental principle in physics that explains how the energy of a system is conserved. Often referred to as the law of energy conservation, it states that energy cannot be created or destroyed in an isolated system. Instead, energy can be transformed from one form to another. For this exercise, remember how heat energy added to the system is converted into internal energy.The law is mathematically expressed as: \[\Delta Q = \Delta U + \Delta W\]where \( \Delta Q \) is the heat added to the system, \( \Delta U \) is the change in internal energy, and \( \Delta W \) is the work done by the system on its surroundings.In scenarios like heating water, if no physical work (e.g., expansion or compression) is done on the surroundings, \( \Delta W = 0 \). Hence, all the heat energy goes into changing the system's internal energy. This results in a simple relation: \( \Delta Q = \Delta U \). This means all the heat energy absorbed by the water contributes directly to increasing its internal energy.
Internal Energy Change
Internal energy is the total energy contained within a system. It involves the movement and interactions of particles, which mean kinetic and potential energy at the molecular level. When you heat a substance like water, energy is added in the form of heat, increasing the water's internal energy.To calculate the change in internal energy (\( \Delta U \)) when heating water, you use the formula:\[\Delta U = c \cdot m \cdot \Delta T\]Here:
  • \( c \) is the specific heat capacity, which measures how much energy is required to raise the temperature of a unit mass by one degree Kelvin.
  • \( m \) is the mass of the substance, in this case, water.
  • \( \Delta T \) is the change in temperature from the initial to the final state.
By applying this equation, the specific heat capacity of water, and knowing the mass and temperature change, you can determine the change in water's internal energy. This reflects the principle of how absorbing or releasing heat affects a substance's internal energy state.
Temperature Conversion
Temperature conversion, especially between Celsius and Kelvin, is an essential step in thermodynamics calculations. The Kelvin scale is often used in scientific context because it is an absolute temperature scale.When dealing with temperature changes like \( \Delta T \), the same change applies whether you're using Celsius or Kelvin. This is because the size of one degree on both the Celsius and Kelvin scales are the same. For example, a temperature change from \( 21^{\circ} \text{C} \) to \( 37^{\circ} \text{C} \) results in \( \Delta T = 16 \) degrees, which is the same in Kelvin.It is crucial when performing calculations involving specific heat and energy change to convert temperatures into Kelvin when dealing with absolute values. Remember that Kelvin is \( 273.15 \) degrees greater than Celsius when converting presence of actual temperatures (e.g., \( T_{\text{K}} = T_{\text{C}} + 273.15 \)). However, for changes \( \Delta T \), the conversion isn't necessary since the change in magnitude is the same.

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Most popular questions from this chapter

A \(2.0\) 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?

A sealed chamber containing \(32.5\) g of molecular oxygen and \(20.2\) g of molecular nitrogen at \(48.0^{\circ} \mathrm{C}\) is cooled down to \(20.2^{\circ} \mathrm{C}\). Given that for \(\mathrm{N}_{2}, c_{v}=0.743 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\), and for \(\mathrm{O}_{2}, c_{v}=0.659\) \(\mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\), determine the resulting change in the internal energy of the gas.

In each of the following situations, find the change in internal energy of the system. (a) A system absorbs 500 cal of heat and at the same time does \(400 \mathrm{~J}\) of work. \((b)\) A system absorbs 300 cal and at the same time \(420 \mathrm{~J}\) of work is done on it. ( \(c\) ) Twelve hundred calories are removed from a gas held at constant volume. Give your answers in kilojoules. (a) \(\Delta U=\Delta Q-\Delta W=(500 \mathrm{cal})(4.184 \mathrm{~J} / \mathrm{cal})-400 \mathrm{~J}=1.69 \mathrm{~kJ}\) (b) \(\Delta U=\Delta Q-\Delta W=(300\) cal \()(4.184 \mathrm{~J} / \mathrm{cal})-(-420 \mathrm{~J})=1.68 \mathrm{~kJ}\) (c) \(\Delta U=\Delta Q-\Delta W=(-1200\) cal \()(4.184 \mathrm{~J} / \mathrm{cal})-0=-5.02 \mathrm{~kJ}\) Notice that \(\Delta Q\) is positive when heat is added to the system and \(\Delta W\) is positive when the system does work. In the reverse cases, \(\Delta Q\) and \(\Delta W\) must be taken negative.

Water is boiled at \(100^{\circ} \mathrm{C}\) and \(1.0\) atm. Under these conditions, 1.0 g of water occupies \(1.0 \mathrm{~cm}^{3}, 1.0\) g of steam occupies 1670 \(\mathrm{cm}^{3}\), and \(L_{v}=540 \mathrm{cal} / \mathrm{g}\). Find \((a)\) the external work done when \(1.0\) g of steam is formed at \(100^{\circ} \mathrm{C}\) and \((b)\) the increase in internal energy.

Twenty cubic centimeters of monatomic gas at \(12{ }^{\circ} \mathrm{C}\) and 100 \(\mathrm{kPa}\) is suddenly (and adiabatically) compressed to \(0.50 \mathrm{~cm}^{3}\). Assume that we are dealing with an ideal gas. What are its new pressure and temperature? For an adiabatic change involving an ideal gas, \(P_{1} V_{1}^{\gamma}=P_{2} V_{2}^{\gamma}\) where \(\mathrm{Y}=1.67\) for a monatomic gas. Hence, $$P_{2}=P_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma}=\left(1.00 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(\frac{20}{0.50}\right)^{1.67}=4.74 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}=47 \mathrm{MPa}$$ To find the final temperature, we could use \(P_{1} V_{1} / T_{1}=P_{2} V_{2} / T_{2}\). Instead, let us use $$T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma}$$ or $$T_{2}=T_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma-1}=(285 \mathrm{~K})\left(\frac{20}{0.50}\right)^{0.67}=(285 \mathrm{~K})(11.8)=3.4 \times 10^{3} \mathrm{~K}$$ As a check, $$\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$$ $$ \begin{aligned} \frac{\left(1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(20 \mathrm{~cm}^{3}\right)}{285 \mathrm{~K}} &=\frac{\left(4.74 \times 107 \mathrm{~N} / \mathrm{m}^{2}\right)\left(0.50 \mathrm{~cm}^{3}\right)}{3370 \mathrm{~K}} \\ 7000 &=7000 \end{aligned} $$

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