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Chapter 3: Problem 45

A well-insulated copper tank of mass \(13 \mathrm{~kg}\) contains \(4 \mathrm{~kg}\) of liquid water. Initially, the temperature of the copper is \(27^{\circ} \mathrm{C}\) and the temperature of the water is \(50^{\circ} \mathrm{C}\). An electrical resistor of neglible mass transfers \(100 \mathrm{~kJ}\) of energy to the contents of the tank. The tank and its contents come to equilibrium. What is the final temperature, in \({ }^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The final temperature is approximately 49.29掳C.

Step by step solution

01

- Understand the Heat Transfer

Energy is being transferred to both the copper tank and the water, raising their temperatures until they reach equilibrium.
02

- Define Variables and Given Information

Let the final temperature be denoted as T. The mass of copper (m_c) is 13 kg, mass of water (m_w) is 4 kg, initial temperature of copper (T_c0) is 27掳C, initial temperature of water (T_w0) is 50掳C, and the energy added (Q) is 100 kJ.
03

- Use Specific Heat Capacities

The specific heat capacity of copper (c_c) is 0.385 kJ/kg掳C and for water (c_w) it is 4.18 kJ/kg掳C.
04

- Set Up Energy Balance Equation

The total energy added to the system is equal to the sum of the energy required to heat the copper and the water to the final temperature: Q = m_c * c_c * (T - T_c0) + m_w * c_w * (T - T_w0).
05

- Plug In Known Values

100 kJ = 13 kg * 0.385 kJ/kg掳C * (T - 27掳C) + 4 kg * 4.18 kJ/kg掳C * (T - 50掳C).
06

- Simplify and Solve for T

Simplifying the equation: 100 = 13 * 0.385 * (T - 27) + 4 * 4.18 * (T - 50) 100 = 5.005 * (T - 27) + 16.72 * (T - 50). Expand and simplify the equation: 100 = 5.005T - 135.135 + 16.72T - 836. Combine like terms: 100 = 21.725T - 971.135. Solve for T: 21.725T = 1071.135. T 鈮� 49.29掳C.

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Key Concepts

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

Energy Balance
In thermodynamics, when we talk about energy balance, we are referring to the conservation of energy principle. This principle states that the total energy in a system remains constant unless there is an exchange with the surroundings.
In our exercise, electrical energy is introduced into the system consisting of a copper tank and water. This energy is distributed between the two, increasing their temperatures until they reach a common temperature, known as thermal equilibrium. The energy balance equation helps in quantifying the distribution and final temperature.
  • Conservation of energy ensures that the sum of energies in each part equals the total added energy.
Specific Heat Capacity
The specific heat capacity of a substance is the amount of heat energy required to raise the temperature of one kilogram of the substance by one degree Celsius. It is a crucial aspect when calculating heat transfer.
In the provided exercise, the specific heat capacities are given for copper and water:
  • For copper: 0.385 kJ/kg掳C
  • For water: 4.18 kJ/kg掳C
These values tell us that water requires more energy per kilogram to change temperature compared to copper. When performing calculations, we multiply these specific heat capacities by the respective masses and temperature changes to understand how much energy each component absorbs.
Thermal Equilibrium
Thermal equilibrium occurs when two objects in thermal contact no longer transfer heat between each other. They attain the same temperature.
In our problem, initially, the copper tank and water are at different temperatures (27掳C for copper and 50掳C for water). Adding 100 kJ of energy causes them to adjust their temperatures until they are the same, achieving thermal equilibrium. This final temperature can be derived through the energy balance equation.
Temperature Change Calculations
Calculating temperature changes involves applying the formula rooted in specific heat capacity:
\[Q = m \times c \times \Delta T\]
Here, Q is the heat added, m is the mass, c is the specific heat capacity, and \Delta T is the change in temperature.
  • Identify mass and specific heat for each material.
  • Set up the equation such that the total energy added equals sum of energies absorbed by both materials.
  • Solve for the final temperature, T.
Following these steps allows us to find the final temperature accurately, considering how much energy each part of the system absorbs and distributes, leading to the equilibrium temperature.

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