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Chapter 18: Problem 1

(a) How much heat is required to raise the temperature of \(250 \mathrm{~mL}\) of water from \(20.0^{\circ} \mathrm{C}\) to \(35.0^{\circ} \mathrm{C}\) ? (b) How much heat is lost by the water as it cools back down to \(20.0{ }^{\circ} \mathrm{C} ?\) Since \(250 \mathrm{~mL}\) of water has a mass of \(250 \mathrm{~g}\), and since \(c=1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) for water, we have (a) \(\Delta Q=m c \Delta T=(250 \mathrm{~g})\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(15.0{ }^{\circ} \mathrm{C}\right)=3.75 \times 10^{3} \mathrm{cal}=15.7 \mathrm{~kJ}\) (b) \(\Delta Q=m c \Delta T=(250 \mathrm{~g})\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(-15.0{ }^{\circ} \mathrm{C}\right)=-3.75 \times 10^{3} \mathrm{cal}=-15.7 \mathrm{~kJ}\) Notice that heat-in (i.e., the heat that enters an object) is taken to be positive, whereas heat-out (i.e., the heat that leaves an object) is taken to be negative.

Short Answer

Expert verified
(a) 15.7 kJ is required; (b) 15.7 kJ is lost.

Step by step solution

01

Understand Given Values and Formulas

The problem provides that the mass of water is 250 g and the specific heat capacity of water is 1.00 cal/(g路掳C). We need to calculate the heat required to change the temperature and can use the formula \( \Delta Q = mc\Delta T \), where \( \Delta Q \) is the heat added or lost, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.
02

Calculate Temperature Change

Determine the temperature change as water heats up and cools down. \( \Delta T = 35.0^{\circ} \mathrm{C} - 20.0^{\circ} \mathrm{C} = 15.0^{\circ} \mathrm{C} \) for heating and \( \Delta T = 20.0^{\circ} \mathrm{C} - 35.0^{\circ} \mathrm{C} = -15.0^{\circ} \mathrm{C} \) when cooling down.
03

Heat Required for Heating (Part a)

Substitute the values into the heat formula to find the heat required to raise the temperature: \( \Delta Q = (250 \mathrm{~g})(1.00 \mathrm{cal}/\mathrm{g} \cdot^{\circ} \mathrm{C})(15.0^{\circ} \mathrm{C}) = 3.75 \times 10^{3} \mathrm{cal} \). Convert calories to kilojoules: \( 3.75 \times 10^3 \mathrm{cal} = 15.7 \mathrm{~kJ} \) (since 1 cal = 4.184 J).
04

Heat Lost during Cooling (Part b)

Use the same formula to calculate heat lost, keeping in mind heat lost is given a negative sign: \( \Delta Q = (250 \mathrm{~g})(1.00 \mathrm{cal}/\mathrm{g} \cdot^{\circ} \mathrm{C})(-15.0^{\circ} \mathrm{C}) = -3.75 \times 10^3 \mathrm{cal} \). Convert negative calories to kilojoules: \( -3.75 \times 10^3 \mathrm{cal} = -15.7 \mathrm{~kJ} \).
05

Analyze Signs of Heat Calculations

By convention, positive heat indicates heat added to a system, whereas negative heat indicates heat removed. Therefore, \( 15.7 \mathrm{~kJ} \) is required to raise the temperature of the water, and \( -15.7 \mathrm{~kJ} \) is lost when it cools.

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Key Concepts

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

Heat Transfer
When we talk about heat transfer, we're looking at the energy moving from one place or substance to another. Heat flows from warmer objects to cooler ones. It's like how when you hold a cold drink, your hand feels cold because warmth from your hand is transferring to the drink to warm it up.
In the given exercise, heat transfer is observed when heat is required to raise the temperature of water and when heat is lost as water cools. This process is governed by the formula for heat transfer:
  • \[ \Delta Q = mc\Delta T \]
Here, \( \Delta Q \) represents the amount of heat transferred, \( m \) is the mass of the object, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature. Understanding this concept helps us get a clear picture of how energy works in our everyday activities, from warming our water to cooling down a hot drink.
Temperature Change
Temperature change is a significant concept when studying heat transfer. It's the difference in temperature that prompts heat to transfer.
In the exercise, the water's temperature moves from 20.0掳C to 35.0掳C and back down again. We calculate temperature change using the formula:
  • \[ \Delta T = T_{ ext{final}} - T_{ ext{initial}} \]
In this formula, \( T_{\text{final}} \) is the temperature you end up with, and \( T_{\text{initial}} \) is the temperature you start with.
After calculation, we see the temperature changes by +15.0掳C when heated and by -15.0掳C when it cools. Positive and negative signs indicate whether the temperature is rising or falling. Knowing how to calculate these changes helps you determine how much energy is involved in warming or cooling an object.
Calorimetry
Calorimetry is the science of measuring heat changes during physical or chemical processes. It's like a detective tool for finding out how much heat energy is transferred. In this exercise, calorimetry helps us calculate the heat required to warm the water and the heat lost when the water cools.

Using the formula \( \Delta Q = mc\Delta T \), calorimetry allows us to measure energy changes:
  • Positive \( \Delta Q \) indicates heat gained by the system.
  • Negative \( \Delta Q \) shows heat lost by the system.
Through calorimetry, we understand that 15.7 kJ of energy is needed to heat the water, and the same amount of energy, but negative, is released when the water cools down. This method aids in various scientific fields, from finding out how much energy is in food to constructing heating systems. Understanding calorimetry empowers us to better interpret everyday energy exchanges and solve related problems.

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Most popular questions from this chapter

Determine the result when \(10 \mathrm{~g}\) of steam at \(100{ }^{\circ} \mathrm{C}\) is passed into a mixture of \(400 \mathrm{~g}\) of water and \(100 \mathrm{~g}\) of ice at exactly \(0{ }^{\circ} \mathrm{C}\) in a calorimeter which behaves thermally as if it were equivalent to \(50 \mathrm{~g}\) of water.

A \(20-\mathrm{g}\) piece of aluminum \(\left(c=0.21 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) at \(90^{\circ} \mathrm{C}\) is dropped into a cavity in a large block of ice at \(0{ }^{\circ} \mathrm{C}\). How much ice does the aluminum melt? (Heat change of Al as it cools to \(0{ }^{\circ} \mathrm{C}\) ) \(+\) (Heat change of mass \(m\) of ice melted) \(=0\) $$ \begin{array}{r} (m c \Delta T)_{\mathrm{Al}}+\left(L_{f} m\right)_{\mathrm{ice}}=0 \\ (20 \mathrm{~g})\left(0.21 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\left(0{ }^{\circ} \mathrm{C}-90^{\circ} \mathrm{C}\right)+(80 \mathrm{cal} / \mathrm{g}) m=0 \end{array} $$ from which \(m=4.7 \mathrm{~g}\) is the quantity of ice melted.

A thermos bottle contains \(250 \mathrm{~g}\) of coffee at \(90{ }^{\circ} \mathrm{C}\). To this is added \(20 \mathrm{~g}\) of milk at \(5{ }^{\circ} \mathrm{C}\). After equilibrium is established, what is the temperature of the liquid? Assume no heat loss to the thermos bottle. Water, coffee, and milk all have the same value of \(c, 1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). The law of energy conservation allows us to write (Heat change of coffee) \(+\) (Heat change of milk) \(=0\) $$ (\mathrm{cm} \Delta T)_{\text {coffee }}+(\mathrm{cm} \Delta T)_{\text {milk }}=0 $$ In other words, the heat lost by the coffee equals the heat gained by the milk. If the final temperature of the liquid is \(T_{f}\), then $$ \Delta T_{\text {coffee }}=T_{f}=90^{\circ} \mathrm{C} \quad \Delta T_{\text {milk }}=T_{f}-5^{\circ} \mathrm{C} $$ Substituting and canceling \(c\) yields $$ (250 \mathrm{~g})\left(T_{f}-90^{\circ} \mathrm{C}\right)+(20 \mathrm{~g})\left(T_{f}-5{ }^{\circ} \mathrm{C}\right)=0 $$ Solving this equation leads to \(T_{f}=84{ }^{\circ} \mathrm{C}\).

Determine the temperature \(T_{f}\) that results when \(150 \mathrm{~g}\) of ice at \(0{ }^{\circ} \mathrm{C}\) is mixed with \(300 \mathrm{~g}\) of water at \(50^{\circ} \mathrm{C}\). From energy conservation, $$ \begin{array}{r} \text { (Heat change of ice) }+\text { (Heat change of water) }=0 \\ \text { (Heat to melt ice) }+\text { (Heat to warm ice water) }+(\text { Heat change of water })=0 \\ \left(m L_{f \text { ice }}+\left(\text { (cm } \Delta T_{\text {ice water }}+(c m \Delta T)_{\text {water }}=0\right.\right. \\ (150 \mathrm{~g})(80 \mathrm{cal} / \mathrm{g})+\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(150 \mathrm{~g})\left(T_{f}-0^{\circ} \mathrm{C}\right)+\left(1.00 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(300 \mathrm{~g})\left(T_{f}-50^{\circ} \mathrm{C}\right)=0 \end{array} $$ from which \(T_{f}=6.7^{\circ} \mathrm{C}\)

Two identical metal plates (mass \(=m\), specific heat \(=c\) ) have different temperatures; one is at \(20{ }^{\circ} \mathrm{C}\), and the other is at \(90^{\circ} \mathrm{C}\). They are placed in good thermal contact. What is their final temperature? Because the plates are identical, we would guess the final temperature to be midway between \(20^{\circ} \mathrm{C}\) and \(90^{\circ} \mathrm{C}\), namely \(55^{\circ} \mathrm{C}\). This is correct, but let us show it mathematically. From the law of conservation of energy, the heat lost by one plate must equal the heat gained by the other. Thus, the total heat change of the system is zero. In equation form, (Heat change of hot plate) \(+(\) Heat change of cold plate \()=0\) $$ m c(\Delta T)_{\mathrm{hot}}+m c(\Delta T)_{\mathrm{cold}}=0 $$ which is short-hand for \(m_{\text {hot }} c_{\text {hot }} \Delta T_{\text {hot }}+m_{\text {cold }} c_{\text {cold }} \Delta T_{\text {cold }}=0\). Be careful about \(\Delta T\) : It is the final temperature (which we denote by \(T_{f}\) in this case) minus the initial temperature. The above equation thus becomes $$ m c\left(T_{f}-90{ }^{\circ} \mathrm{C}\right)+m c\left(T_{f}-20{ }^{\circ} \mathrm{C}\right)=0 $$ After canceling \(m c\) from each term, solve the equation and find \(T_{f}=55^{\circ} \mathrm{C}\), the expected answer.
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