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Chapter 9: Problem 43

Assume that the kinetic energy of a \(1400-\mathrm{kg}\) car moving at \(115 \mathrm{~km} / \mathrm{h}\) (Problem \(9.42\) ) could be converted entirely into heat. What amount of water could be heated from \(20^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) by the car's energy? \(4.184 \mathrm{~J}\) are required to heat \(1 \mathrm{~g}\) of water by \(1{ }^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The car's kinetic energy can heat approximately 5.714 kg of water from 20掳C to 50掳C.

Step by step solution

01

Convert Car Speed to m/s

First, convert the speed of the car from kilometers per hour to meters per second. This allows us to work with standard SI units.Given: \(115\, \text{km/h} = \frac{115 \times 1000}{3600} \, \text{m/s} = 31.94 \, \text{m/s} \approx 32 \, \text{m/s}\).
02

Calculate Kinetic Energy of the Car

The kinetic energy (KE) of an object can be calculated using the formula: \( KE = \frac{1}{2}mv^2 \) where \(m\) is mass and \(v\) is velocity.Given: Mass \(m = 1400\, \text{kg}\) and velocity \(\approx 32 \, \text{m/s}\),\( KE = \frac{1}{2} \times 1400 \times (32)^2 = 716,800 \, \text{Joules}\).
03

Determine Energy Required to Heat Water

The energy required to heat water is given by \( Q = mc\Delta T \), where \(m\) is the mass of water, \(c\) is the specific heat capacity, and \(\Delta T\) is the temperature change.For water, \(c = 4.184 \,\text{J/g}^\circ\text{C}\) and \(\Delta T = 50^\circ\text{C} - 20^\circ\text{C} = 30^\circ\text{C}\).
04

Calculate Mass of Water Heated

Set the kinetic energy of the car equal to the energy needed to heat the water to find the mass \(m\) of the water.\(716,800 = m \times 4.184 \times 30\)\(m = \frac{716,800}{4.184 \times 30} \approx 5,714 \,\text{g}\).Thus, approximately \(5,714\,\text{g}\) or \(5.714\,\text{kg}\) of water can be heated from \(20^\circ\text{C}\) to \(50^\circ\text{C}\).

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Key Concepts

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

Specific Heat Capacity
Specific heat capacity is a fundamental property that measures the heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius. It's an important factor in various scientific fields and everyday applications.

For example, water has a specific heat capacity of 4.184 J/g掳C, which is quite high. This means that water can absorb a lot of heat before its temperature rises significantly.

This property is why water is a great medium for heating and cooling systems.
  • Substances with high specific heat capacity like water can stabilize temperature changes.
  • They require more energy for the same amount of temperature increase compared to substances with lower specific heat capacity.
Understanding and utilizing specific heat capacity helps in designing systems like climate control and energy storage.
Energy Conversion
Energy conversion is the process of changing energy from one form to another. It plays a vital role in various engines and mechanical processes.

In our example, the kinetic energy of a moving car gets converted into heat energy. This transformation illustrates how energy doesn't vanish but just changes its form.

Such conversions are essential to understand when considering systems like car engines and power plants.
  • Kinetic energy can be transformed into electrical, heat, or potential energy.
  • Efficient energy conversion maximizes output and minimizes waste.
In practice, understanding energy conversion allows engineers to create more efficient machines and systems.
Kinetic Energy Formula
Kinetic energy is the energy of motion, and the formula for kinetic energy is beautifully simple: \( KE = \frac{1}{2}mv^2 \). This formula requires knowing the mass (\(m\)) of an object and its velocity (\(v\)).

Kinetic energy depends on both mass and speed. Doubling the speed of an object results in four times more kinetic energy, as the velocity is squared in the formula.

For instance, a heavier or faster-moving car has significantly larger kinetic energy than a lighter or slower one.
  • Mass and velocity both contribute to an object's kinetic energy.
  • The quadratic relationship with speed (velocity squared) highlights the drastic increase as an object moves faster.
Mastering this formula is crucial for solving physics problems involving motion and dynamics.
Temperature Change
Temperature change represents the difference in temperature as heat energy is added or removed from a substance. In our context, it specifically looks at how much an object's temperature rises as it gains energy.

The formula for calculating required energy involves temperature change \(\Delta T\), mass \(m\), and specific heat capacity \(c\): \( Q = mc\Delta T \).

Knowing the initial and final temperatures allows us to find \(\Delta T\), which determines how much energy is needed for that change.
  • Temperature change is a critical factor in heating and cooling processes.
  • Understanding \(\Delta T\) helps in processes like cooking, heating systems, and chemical reactions.
Grasping temperature change concepts ensures accurate calculations for energy requirements in real-world applications.

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Most popular questions from this chapter

How much heat in kilojoules is evolved or absorbed in the reaction of \(2.50 \mathrm{~g}\) of \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) with enough carbon monoxide to produce iron metal? Is the process exothermic or endothermic? $$\begin{aligned}\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{CO}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{CO}_{2}(g) \\ \Delta H^{\circ}=-24.8 \mathrm{~kJ}\end{aligned}$$

Sulfuric acid \(\left(\mathrm{H}_{2} \mathrm{SO}_{4}\right)\), the most widely produced chemical in the world, is made by a two-step oxidation of sulfur to sulfur trioxide, \(\mathrm{SO}_{3}\), followed by reaction with water. Calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{SO}_{3}\) in \(\mathrm{kJ} / \mathrm{mol}\), given the following data: $$\begin{array}{ll}\mathrm{S}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g) & \Delta H^{\circ}=-296.8 \mathrm{~kJ} \\ \mathrm{SO}_{2}(g)+1 / 2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{SO}_{3}(g) & \Delta H^{\circ}=-98.9 \mathrm{~kJ} \end{array}$$

Ethyl alcohol has \(\Delta H_{\text {fusion }}=5.02 \mathrm{~kJ} / \mathrm{mol}\) and melts at \(-114.1^{\circ} \mathrm{C}\). What is the value of \(\Delta S_{\text {fusion }}\) for ethyl alcohol?

When \(25.0 \mathrm{~mL}\) of \(1.0 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) is added to \(50.0 \mathrm{~mL}\) of \(1.0 \mathrm{M} \mathrm{NaOH}\) at \(25.0^{\circ} \mathrm{C}\) in a calorimeter, the temperature of the aqueous solution increases to \(33.9^{\circ} \mathrm{C}\). Assuming that the specific heat of the solution is \(4.18 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\), that its density is \(1.00 \mathrm{~g} / \mathrm{mL}\), and that the calorimeter itself absorbs a negligible amount of heat, calculate \(\Delta H\) in kilojoules \(/ \mathrm{mol}\) \(\mathrm{H}_{2} \mathrm{SO}_{4}\) for the reaction. $$\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{NaOH}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q)$$

We said in Section \(9.1\) that the potential energy of water at the top of a dam or waterfall is converted into heat when the water dashes against rocks at the bottom. The potential energy of the water at the top is equal to \(E_{p}=m g h\), where \(m\) is the mass of the water, \(g\) is the acceleration of the falling water due to gravity \(\left(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\), and \(h\) is the height of the water. Assuming that all the energy is converted to heat, calculate the temperature rise of the water in degrees Celsius after falling over California's Yosemite Falls, a distance of \(739 \mathrm{~m}\). The specific heat of water is \(4.18 \mathrm{~J} /(\mathrm{g} \cdot \mathrm{K})\)
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