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Chapter 2: Problem 5

Count Rumford observed that using cannon boring machinery a single horse could heat \(11.6 \mathrm{kg}\) of ice water \((T=273 \mathrm{K})\) to \(T=355 \mathrm{K}\) in 2.5 hours. Assuming the same rate of work, how high could a horse raise a 225 kg weight in 2.5 minutes? Assume the heat capacity of water is \(4.18 \mathrm{JK}^{-1} \mathrm{g}^{-1}\)

Short Answer

Expert verified
The height to which a horse can raise a 225 kg weight in 2.5 minutes can be calculated using the conservation of energy through the following steps: 1) Calculate the heat absorbed by the water (Q), 2) Calculate the power of the horse, 3) Calculate the work done in raising the weight, and 4) Use the gravitational potential energy formula to find the height. Following these steps, we find that the height the horse can raise the 225 kg weight in 2.5 minutes is given by: \(h = \frac{W}{(225)(9.81)}\)

Step by step solution

01

Calculate the heat absorbed by the water.

The heat absorbed by the water can be calculated using the formula: $$Q = mc(T_2-T_1),$$ where \(Q\) is the heat absorbed, \(m\) is the mass of the water, \(c\) is the specific heat capacity of water, and \(T_2\) and \(T_1\) are the final and initial temperatures, respectively. The mass (\(m\)) of the water is given by 11.6 kg, the specific heat capacity (\(c\)) is given as 4.18 J/(gK), which we need to convert to J/(kgK), so multiply by 1000: \(4.18 \times 1000 = 4180 \: \text{J}\;(K\cdot \text{kg})^{-1}\). The temperatures are given as \(T_1 = 273 \: \text{K}\) and \(T_2 = 355 \: \text{K}\). Using the given values, let's now calculate the heat absorbed by the water: \(Q = (11.6) \times (4180) \times (355-273)\)
02

Calculate the power of the horse.

The power (\(P\)) of the horse can be calculated as the amount of heat absorbed by the water in 2.5 hours. Since the heat was absorbed over 2.5 hours, we need to convert the time to seconds: \(2.5 \: \text{hrs} \times 60 \: \text{min}/\text{hr} \times 60 \: \text{sec}/\text{min} = 9000 \: \text{sec}\). Now, we can calculate the power of the horse using the formula: Power = Work/Time \(P = \frac{Q}{t}\), where \(t\) is the time (in seconds) taken for the work to be done. Using the values from step 1, we calculate the power of the horse: \(P = \frac{Q}{9000}\)
03

Calculate the work done in raising the weight.

The horse raises the weight in 2.5 minutes, so first convert the time to seconds: \(2.5 \: \text{min} \times 60 \: \text{sec}/\text{min} = 150 \: \text{sec}\). Using the same power, we can calculate the work done (\(W\)) in raising the weight using the same formula for power: \(W = Pt\). Now, using the value of \(P\) from step 2 and the time of 150 seconds, let's calculate the work done in raising the weight: \(W = P \times (150)\)
04

Calculate the height the weight is raised.

To calculate the height (\(h\)) the weight is raised, we use the formula for gravitational potential energy: \(W = mgh\), where \(m\) is the mass of the weight, and \(g\) is the acceleration due to gravity (approximately \(9.81 \: \text{m/s}^2\)). Now, we can solve for the height \(h\): \(h = \frac{W}{mg}\) Using the mass of the weight as 225 kg, the value of \(g\) as 9.81 m/s虏, and the value of W from step 3, let's calculate the height the horse can raise the weight: \(h = \frac{W}{(225)(9.81)}\) Finally, we have the height the horse can raise the 225 kg weight in 2.5 minutes.

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Key Concepts

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

Heat Capacity
Heat capacity is a fundamental concept in thermodynamics. It refers to the amount of heat energy required to increase the temperature of a substance by a specified amount. In our scenario, we are dealing with water whose heat capacity is known.
The formula used to calculate the heat absorbed is:
  • \( Q = mc(T_2 - T_1) \)
Here, \( Q \) is the heat absorbed, \( m \) is the mass, \( c \) is the specific heat capacity, and \( T_2 - T_1 \) is the change in temperature.
This helps us understand how much energy is transferred into the water to raise its temperature from one state to another in our practical problem.
Power Calculation
Power is the rate at which work is done or energy is transferred. It is measured in watts (W), with one watt equal to one joule per second. In the example provided, we calculated the power of the horse by determining how much heat energy from the horse was absorbed by the water, over a period of time.
The power can be calculated using the formula:
  • \( P = \frac{Q}{t} \)
Here, \( Q \) is the amount of energy in joules, and \( t \) is the time in seconds.
By determining the power, we gain insight into the horse's capacity to perform work in a specific timeframe.
Gravitational Potential Energy
Gravitational potential energy is the energy an object possesses due to its height above the ground. It depends on the mass of the object, the height it is raised to, and the gravitational force. We can express it with the formula:
  • \( W = mgh \)
Using this formula, we find the work done when raising a weight.
Here, \( W \) is the work done, \( m \) is the mass, \( g \) is the gravitational acceleration (approx. \(9.81 \,\text{m/s}^2\)), and \( h \) is the height.
This concept is crucial for solving the problem, as it connects the mechanical work done to raise the weight with the heat absorbed by the water.
Specific Heat
Specific heat is the amount of heat needed to raise the temperature of a unit mass of a substance by one degree Celsius. It plays a crucial role when calculating heat energy changes in thermodynamic processes.
For example, the specific heat capacity of water here is given as \(4.18 \,\text{J/(gK)}\). Since the problem involves a mass in kilograms, we convert it to \(4180 \,\text{J/(kgK)}\) to maintain consistency in units.
By understanding specific heat, we can effectively calculate the heat absorbed by different substances, which is especially relevant when dealing with temperature changes in a quantitative manner.

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Most popular questions from this chapter

A vessel containing 1.50 mol of an ideal gas with \(P_{i}=1.00\) bar and \(C_{P, m}=5 R / 2\) is in thermal contact with a water bath. Treat the vessel, gas, and water bath as being in thermal equilibrium, initially at \(298 \mathrm{K},\) and as separated by adiabatic walls from the rest of the universe. The vessel, gas, and water bath have an average heat capacity of \(C_{P}=2450 . \mathrm{JK}^{-1} .\) The gas is compressed reversibly to \(P_{f}=20.5\) bar. What is the temperature of the system after thermal equilibrium has been established?

A 1.75 mole sample of an ideal gas for which \(C_{V, m}=20.8 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\) is heated from an initial temperature of \(21.2^{\circ} \mathrm{C}\) to a final temperature of \(380 .^{\circ} \mathrm{C}\) at constant volume. Calculate \(q, w, \Delta U,\) and \(\Delta H\) for this process.

A nearly flat bicycle tire becomes noticeably warmer after it has been pumped up. Approximate this process as a reversible adiabatic compression. Assume the initial pressure and temperature of the air before it is put in the tire to be \(P_{i}=1.00\) bar and \(T_{i}=280 .\) K. The final pressure in the tire is \(P_{f}=3.75\) bar. Calculate the final temperature of the air in the tire. Assume that \(C_{V, \pi}=5 R / 2\)

Consider the isothermal expansion of 2.35 mol of an ideal gas at \(415 \mathrm{K}\) from an initial pressure of 18.0 bar to a final pressure of 1.75 bar. Describe the process that will result in the greatest amount of work being done by the system with \(P_{\text {external }} \geq 1.75\) bar, and calculate \(w\). Describe the process that will result in the least amount of work being done by the system with \(P_{\text {external}} \geq 1.75\) bar, and calculate \(w .\) What is the least amount of work done without restrictions on the external pressure?

The formalism of the Young's modulus is sometimes used to calculate the reversible work involved in extending or compressing an elastic material. Assume a force \(F\) is applied to an elastic rod of cross-sectional area \(A_{0}\) and length \(L_{0}\). As a result of this force the rod changes in length by \(\Delta L\). The Young's modulus \(E\) is defined as $$E=\frac{\text { tensile stress }}{\text { tensile strain }}=\frac{F / A_{0}}{\Delta L / L_{0}}=\frac{F L_{0}}{A_{0} \Delta L}$$ a. Relate \(k\) in Hooke's Law to the Young's modulus expression just given. b. Using your result in part (a) show that the magnitude of the reversible work involved in changing the length \(L_{0}\) of an elastic cylinder of cross- sectional area \(A_{0}\) by \(\Delta L\) is \(w=\frac{1}{2}\left(\frac{\Delta L}{L_{0}}\right)^{2} E A_{0} L_{0}\)
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