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Chapter 13: Problem 4

An amount of water of mass \(20 \mathrm{~g}\) at \(0^{\circ} \mathrm{C}\) is mixed with 40 \(\mathrm{g}\) of water at \(10^{\circ} \mathrm{C}\), final temperature of the mixture is (a) \(5^{\circ} \mathrm{C}\) (b) \(0^{\circ} \mathrm{C}\) (c) \(20^{\circ} \mathrm{C}\) (d) \(6.66^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The final temperature is \(6.66^{\circ} \mathrm{C}\), option (d).

Step by step solution

01

Understand the Problem

We need to determine the final temperature when 20 g of water at \(0^{\circ} \mathrm{C}\) is mixed with 40 g of water at \(10^{\circ} \mathrm{C}\). The options given are \(5^{\circ} \mathrm{C}\), \(0^{\circ} \mathrm{C}\), \(20^{\circ} \mathrm{C}\), and \(6.66^{\circ} \mathrm{C}\). This is a heat exchange problem where the heat lost by the warmer water is equal to the heat gained by the colder water.
02

Define the Heat Exchange Equation

Use the formula for heat exchange, which is based on specific heat capacity: \(m_1 c (T - T_1) = m_2 c (T_2 - T)\), where \(m_1\) and \(m_2\) are the masses of the water, \(c\) is the specific heat capacity of water, \(T_1\) and \(T_2\) are initial temperatures, and \(T\) is final equilibrium temperature. For water, \(c = 4.18 \, \text{J/g}^\circ\text{C}\) but it cancels out in the equation.
03

Set Up the Equation with Given Values

Substitute the given values into the heat exchange equation: \(20 \cdot (T - 0) = 40 \cdot (10 - T)\). This simplifies to \(20T = 400 - 40T\).
04

Solve for the Final Temperature

Combine terms: \(20T + 40T = 400\), which simplifies to \(60T = 400\). Divide both sides by 60, which gives \(T = \frac{400}{60} = 6.66\ldots\). So, the final temperature \(T = 6.66^{\circ} \mathrm{C}\).
05

Select the Correct Answer

Compare the calculated temperature with the options provided. The closest match is \(6.66^{\circ} \mathrm{C}\), which is option (d).

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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 an essential concept in understanding how substances absorb and release heat. It is defined as the amount of heat per unit mass required to raise the temperature by one degree Celsius. For water, the specific heat capacity is quite high at 4.18 J/g掳C. This means water can absorb a significant amount of heat without a significant change in temperature, making it a great substance for thermal regulation.
When dealing with heat exchange problems, specific heat capacity allows us to predict how much heat is transferred between substances of different temperatures. In this exercise, both quantities of water (even though different in mass) work with the same specific heat capacity. Thus, it becomes easier to calculate how heat will flow from the warmer to the cooler water until they reach a common temperature. Calculating heat using specific heat capacity allows scientists and engineers to design systems for managing heat in various contexts, from engine cooling to climate control systems.
Temperature Equilibrium
Temperature equilibrium in a system is achieved when two or more substances at different initial temperatures come into contact and exchange heat until they all reach the same final temperature. This final temperature is called the equilibrium temperature.
When mixing two amounts of water at different temperatures, as in this exercise, the heat lost by the warmer water is equal to the heat gained by the cooler water. This is because water follows the law of conservation of energy, meaning energy can neither be created nor destroyed, only transferred.
For processes like this, when the system reaches equilibrium, there is no more net flow of thermal energy between the two water masses. The temperature throughout the system is uniform. Understanding temperature equilibrium is crucial for many scientific applications, including designing thermal systems or even everyday situations like mixing a hot and cold drink.
Mixing Temperature Calculation
The mixing temperature calculation involves using a simple heat exchange formula that aligns with the conservation of energy principle. The formula used in these calculations is derived from equating the heat lost by the hot water to the heat gained by the cold water.
The general equation for this scenario is \(m_1 c (T - T_1) = m_2 c (T_2 - T)\), where:
  • \(m_1\) and \(m_2\) are the masses of the substances (in this case, water)
  • \(T_1\) and \(T_2\) are the initial temperatures of each mass
  • \(T\) is the final equilibrium temperature
Given that the specific heat capacity (\( c \)) of water is consistent, it cancels out, simplifying the calculation. In our example, plugging the values into the formula gives us: \((20)(T - 0) = (40)(10 - T)\). Solving this equation leads to finding the equilibrium temperature as \(6.66^\circ\\) C.
This exercise highlights how temperature changes depend on the relative amounts of water and their starting temperatures.

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Most popular questions from this chapter

Inside a cylinder closed at both ends is a movable piston. On one side of the piston is a mass \(m\) of a gas, and on the other side a mass \(2 m\) of the same gas. What fraction of the volume of the cylinder will be occupied by the larger mass of the gas when the piston is in equilibrium? The temperature is the same throughout. (a) \(\frac{2}{3}\) (b) \(\frac{1}{3}\) (c) \(\frac{1}{2}\) (d) \(\frac{1}{4}\)

Two rods of same length and material transfer a given amount of heat in \(12 \mathrm{~s}\), when they are joined end to end (i.e., in series). But when they are joined in parallel, they will transfer same heat under same conditions in (a) 245 (b) \(3 \underline{5}\) (c) \(48 \mathrm{~s}\) (d) \(1.5 \mathrm{~s}\)

Two rods \(P\) and \(Q\) have equal lengths. Their thermal conductivities are \(K_{1}\) and \(K_{2}\) and cross-sectional areas are \(A_{1}\) and \(A_{2}\). When the temperature at ends of each rod are \(T_{1}\) and \(T_{2}\) respectively, the rate of flow of heat through \(P\) and \(Q\) will be equal, if (a) \(\frac{A_{1}}{A_{2}}=\frac{K_{2}}{K_{1}}\) (b) \(\frac{A_{1}}{A_{2}}=\frac{K_{2}}{K_{1}} \times \frac{T_{2}}{T_{1}}\) (c) \(\frac{A_{1}}{A_{2}}=\sqrt{\frac{K_{1}}{K_{2}}}\) (d) \(\frac{A_{1}}{A_{2}}=\left(\frac{K_{1}}{K_{1}}\right)^{2}\)

A wheel is \(80.3 \mathrm{~cm}\) in circumference. An iron tyre measures \(80.0 \mathrm{~cm}\) around its inner face. If the coefficient of linear expansion for iron is \(1.2 \times 0^{-5}{ }^{\circ} \mathrm{C}^{-1}\), the temperature of the tyre must be raised by (a) \(105^{\circ} \mathrm{C}\) (b) \(417^{\circ} \mathrm{C}\) (c) \(312^{+} \mathrm{C}\) (d) \(223^{\circ} \mathrm{C}\)

A substance of mass \(m \mathrm{~kg}\) requires a power input of \(P\) watts to remain in the molten state at its melting point. When the power is turned off, the sample completely solidifies in time \(t\) sec. What is the latent heat of fusion of the substance? (a) \(\frac{P m}{t}\) (b) \(\frac{P t}{m}\) (c) \(\frac{m}{P t}\) (d) \(\frac{t}{P m}\)
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