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A sheet of gold weighing \(10.0 \mathrm{~g}\) and at a temperature of \(18.0^{\circ} \mathrm{C}\) is placed flat on a sheet of iron weighing \(20.0 \mathrm{~g}\) and at a temperature of \(55.6^{\circ} \mathrm{C}\). What is the final temperature of the combined metals? Assume that no heat is lost to the surroundings.

Short Answer

Expert verified
The final temperature of the combined metals is approximately \(50.9^{\circ}C\).

Step by step solution

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01

Understanding the Heat Exchange

In this scenario, since no heat is lost to the surroundings, the heat lost by the sheet of iron will be equal to the heat gained by the sheet of gold. This can be formulated as: \[ q_{ ext{gold}} + q_{ ext{iron}} = 0 \] where \( q \) represents heat.
02

Apply the Heat Transfer Formula

The heat transfer is calculated using the formula: \[ q = mc\Delta T \] where \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature \((T_{ ext{final}} - T_{ ext{initial}})\).
03

Setting Up the Heat Balance Equation

We know the specific heat capacities: - Gold: \( c_{ ext{gold}} = 0.129 \, \text{J/g} \, ^\circ\text{C} \) - Iron: \( c_{ ext{iron}} = 0.450 \, \text{J/g} \, ^\circ\text{C} \) So, the equation becomes: \[ m_{ ext{gold}} c_{ ext{gold}}(T_{ ext{final}} - T_{ ext{gold,initial}}) + m_{ ext{iron}} c_{ ext{iron}}(T_{ ext{final}} - T_{ ext{iron,initial}}) = 0 \] Substituting the values, we get: \[ 10.0 \times 0.129 (T_{ ext{final}} - 18.0) + 20.0 \times 0.450 (T_{ ext{final}} - 55.6) = 0 \]
04

Solve the Equation

First, simplify the equation: \[ 1.29(T_{ ext{final}} - 18.0) + 9.0(T_{ ext{final}} - 55.6) = 0 \] Expanding leads to: \[ 1.29T_{ ext{final}} - 23.22 + 9.0T_{ ext{final}} - 500.4 = 0 \] Combine like terms: \[ 10.29T_{ ext{final}} - 523.62 = 0 \] Solve for \( T_{ ext{final}} \): \[ 10.29T_{ ext{final}} = 523.62 \] \[ T_{ ext{final}} = \frac{523.62}{10.29} \approx 50.9 \]
05

Conclusion: Determine Final Temperature

From the calculation, the final temperature when both metals reach thermal equilibrium is approximately \( 50.9^{ ext{circ}} C \).

Key Concepts

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

Specific Heat Capacity
Every material requires a certain amount of energy to change its temperature. This is where the concept of specific heat capacity comes in. Specific heat capacity, often denoted as 'c,' is the amount of heat needed to increase the temperature of 1 gram of a substance by 1 degree Celsius.
For instance, gold has a specific heat capacity of 0.129 J/g °C, while iron's is higher at 0.450 J/g °C. This means that iron requires more heat for the same mass to achieve the same temperature change compared to gold.
In our exercise, understanding specific heat capacity helps us calculate how much heat each metal gains or loses as they reach a state of equilibrium.
Thermal Equilibrium
Thermal equilibrium is a key principle in heat transfer. It occurs when two or more bodies, in contact, reach the same temperature and no longer exchange heat.
Think of it as the point where they 'agree' on one, final temperature.
In the exercise, the gold and iron sheets reach thermal equilibrium, which allows us to calculate their shared final temperature by balancing the heat lost and gained between them.
  • The heat lost by the warmer iron is exactly equal to the heat gained by the colder gold.
  • This balance is fundamental in finding the final temperature.
Temperature Change
Temperature change is simply the difference between the final temperature and the initial temperature of a substance. It's represented as ΔT in equations.In heat transfer problems like the exercise, ΔT plays a vital role in determining how much heat is exchanged between the objects.
For each metal:
  • The formula is expressed as \( \Delta T = T_{\text{final}} - T_{\text{initial}} \)
  • For gold, the initial temperature is \(18.0\, ^\circ C\), while iron is \(55.6\, ^\circ C\).
This change helps calculate how much heat is required to reach equilibrium.
Heat Exchange
Heat exchange is the process by which heat energy is transferred between substances or systems. In the scenario with gold and iron, the heat lost by the iron and gained by the gold can be expressed with the formula: \[ q = mc\Delta T \]Here:
  • \( q \) is the heat exchanged;
  • \( m \) is mass;
  • \( c \) is specific heat capacity;
  • \( \Delta T \) is the temperature change.
This equation is used to setup a balance between heat lost by iron and heat gained by gold.
A crucial step is setting up the equation such that all exchanged heat sums to zero. This principle ensures energy is conserved, and by solving the upcoming equation, we can find how the final temperature of the two metals is determined.

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Most popular questions from this chapter

(a) A person drinks four glasses of cold water \(\left(3.0^{\circ} \mathrm{C}\right)\) every day. The volume of each glass is \(2.5 \times 10^{2} \mathrm{~mL}\). How much heat (in kJ) does the body have to supply to raise the temperature of the water to \(37^{\circ} \mathrm{C},\) the body temperature? (b) How much heat would your body lose if you were to ingest \(8.0 \times 10^{2} \mathrm{~g}\) of snow at \(0^{\circ} \mathrm{C}\) to quench your thirst? (The amount of heat necessary to melt snow is \(6.01 \mathrm{~kJ} / \mathrm{mol}\).)

You are given the following data: \(\begin{aligned} \mathrm{H}_{2}(g) & \longrightarrow 2 \mathrm{H}(g) & & \Delta H^{\circ}=436.4 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{Br}_{2}(g) & \longrightarrow 2 \mathrm{Br}(g) & & \Delta H^{\circ}=192.5 \mathrm{~kJ} / \mathrm{mol} \\\ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) & \longrightarrow 2 \mathrm{HBr}(g) & \Delta H^{\circ} &=-72.4 \mathrm{~kJ} / \mathrm{mol} \end{aligned}\) Calculate \(\Delta H^{\circ}\) for the reaction\(\mathrm{H}(g)+\mathrm{Br}(g) \longrightarrow \mathrm{HBr}(g)\).

Given the thermochemical data,$$\begin{array}{ll}\mathrm{A}+6 \mathrm{~B} \longrightarrow 4 \mathrm{C} & \Delta H_{1}=-1200 \mathrm{~kJ} / \mathrm{mol} \\\\\mathrm{C}+\mathrm{B} \longrightarrow \mathrm{D} & \Delta H_{1}=-150 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ Determine the enthalpy change for each of the following: a) \(\mathrm{D} \longrightarrow \mathrm{C}+\mathrm{B}\) d) \(2 \mathrm{D} \longrightarrow 2 \mathrm{C}+2 \mathrm{~B}\) b) \(2 \mathrm{C} \longrightarrow \frac{1}{2} \mathrm{~A}+3 \mathrm{~B}\) e) \(6 \mathrm{D}+\mathrm{A} \longrightarrow 10 \mathrm{C}\) c) \(3 \mathrm{D}+\frac{1}{2} \mathrm{~A} \stackrel{\longrightarrow}{\longrightarrow} \mathrm{C}\)

Why are cold, damp air and hot, humid air more uncomfortable than dry air at the same temperatures? [The specific heats of water vapor and air are approximately \(1.9 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) and \(1.0 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\) respectively.

A \(0.1375-\mathrm{g}\) sample of solid magnesium is burned in a constant-volume bomb calorimeter that has a heat capacity of \(3024 \mathrm{~J} /{ }^{\circ} \mathrm{C}\). The temperature increases by \(1.126^{\circ} \mathrm{C}\). Calculate the heat given off by the burning \(\mathrm{Mg},\) in \(\mathrm{kJ} / \mathrm{g}\) and in \(\mathrm{kJ} / \mathrm{mol} .\)

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