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Chapter 14: Problem 54

A \(50.0\) -gram sample of a metal alloy at \(25.0^{\circ} \mathrm{C}\) is placed in \(100.0\) grams of water at \(55.0^{\circ} \mathrm{C}\). The final temperature of the water and metal is \(48.1^{\circ} \mathrm{C}\). Calculate the specific heat of the alloy.

Short Answer

Expert verified
The specific heat of the alloy is approximately 2.50 J/g掳C.

Step by step solution

01

Understand the System

When the metal alloy and water are mixed, heat will flow from the water to the metal until they reach thermal equilibrium at the same final temperature. The heat lost by the water is equal to the heat gained by the metal.
02

Apply the Formula for Heat Transfer

Consider the formula for heat transfer: \[q = m \cdot c \cdot \Delta T\] where \(q\) is the heat energy, \(m\) is mass, \(c\) is specific heat capacity, and \(\Delta T\) is the change in temperature. Heat lost by water will be equal to the negative heat gained by the alloy. Thus, \[-q_{water} = q_{alloy}\].
03

Calculate Water's Heat Loss

Use the formula for heat to find the heat loss of water: \[q_{water} = m_{water} \cdot c_{water} \cdot (T_{final} - T_{initial, water})\]Given: \(m_{water} = 100.0 \text{ g}, c_{water} = 4.18 \text{ J/g掳C}, T_{initial, water} = 55.0\,^{\circ}\text{C}, T_{final} = 48.1\,^{\circ}\text{C}\). Calculate:\[q_{water} = 100.0 \cdot 4.18 \cdot (48.1 - 55.0)\]\[q_{water} = 100.0 \cdot 4.18 \cdot (-6.9)\]\[q_{water} = -2884.2 \text{ J}\].
04

Equate Heat Exchanges

Use the equation: \[-q_{water} = q_{alloy}\]. Thus, the heat gained by the metal alloy is \[q_{alloy} = 2884.2 \text{ J}\].
05

Solve for Specific Heat of Alloy

Use the heat equation for the metal: \[q_{metal} = m_{metal} \cdot c_{metal} \cdot (T_{final} - T_{initial, metal})\]Substitute in the known values:\[2884.2 = 50.0 \cdot c_{metal} \cdot (48.1 - 25.0)\]Solve for \(c_{metal}\):\[2884.2 = 50.0 \cdot c_{metal} \cdot 23.1\]\[c_{metal} = \frac{2884.2}{50.0 \cdot 23.1}\]\[c_{metal} = \frac{2884.2}{1155.0}\]\[c_{metal} \approx 2.50 \text{ J/g掳C}\].

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Key Concepts

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

Heat Transfer
Heat transfer is a fundamental concept in thermodynamics, involving the movement of thermal energy from one object or substance to another. This movement occurs due to a temperature difference between the substances. In the original problem, heat is transferred from the hotter water to the cooler metal alloy until they both reach a common temperature. This process is governed by the principle of conservation of energy.

Here are some key points about heat transfer:
  • Heat flows from the warmer object to the cooler one until equilibrium is established.
  • The amount of heat transferred depends on the mass, specific heat capacity, and temperature change of the substances involved.
  • Mathematically, heat transfer is represented by: \[q = m \cdot c \cdot \Delta T\] where \(q\) is heat, \(m\) is mass, \(c\) is specific heat capacity, and \(\Delta T\) is the change in temperature.
Understanding heat transfer helps explain everyday phenomena, such as why a spoon heats up in a cup of hot coffee and why a seatbelt buckle gets warm on a sunny day.
Thermal Equilibrium
Thermal equilibrium occurs when two objects in contact with each other reach the same temperature, and therefore, no more heat is exchanged between them. In the context of the exercise, it refers to the situation where the metal alloy and water reach a uniform temperature of \(48.1^{\circ} \text{C}\).

Some points to consider about thermal equilibrium:
  • Once equilibrium is reached, the temperatures of the two substances will remain constant if no further external heat sources are introduced.
  • The heat energy lost by the warmer water becomes equal to the heat gained by the cooler metal, ensuring no net heat flow.
  • This concept allows us to solve for unknowns, like the specific heat of a metal, because we can apply the principle of equal and opposite heat transfer: \[-q_{\text{water}} = q_{\text{metal}} \].
Thermal equilibrium is a vital principle for understanding how heat dissipates in systems, helping everything from engineering applications to climate science.
Specific Heat Capacity
Specific heat capacity is a property of a material that describes how much heat is required to change the temperature of a unit mass of the material by one degree Celsius. It is a crucial concept for understanding how different substances respond to thermal energy, and in the exercise, you calculated the specific heat capacity of a metal alloy.

Here are some important aspects of specific heat capacity:
  • Specific heat capacity is denoted by the symbol \(c\) and has units of \(\text{J/g掳C}\).
  • Substances with high specific heat capacities require more energy to change their temperature, such as water, which moderates climate due to its high specific heat.
  • During calculations like in the original exercise, after equating the heat transfer quantities, one can solve for specific heat capacity using: \[c = \frac{q}{m \cdot \Delta T}\] where \(q\) is the heat gained or lost, \(m\) is mass, and \(\Delta T\) is the change in temperature.
By knowing the specific heat capacity of materials, we better understand how different objects will heat up or cool down, leading to various practical applications in cooking, industry, and even consumer product design.

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Most popular questions from this chapter

One of the most famous equations in science is \(E=m c^{2}\), which comes from Einstein's theory of relativity. Einstein was the first to show that mass can be converted into energy, and that energy can be converted into mass. This equation relates the amount of energy produced when the mass of a system decreases, or conversely, the amount of energy that is required to increase the mass of a system. In either case, the equation relates the change in energy associated with a change in mass, and in the notation of this chapter, Einstein's mass-energy relation is written as $$ \Delta U=c^{2} \Delta m $$ where \(\Delta U\) is the change in energy and \(\Delta m\) is the change in mass. Calculate the amount of energy produced when a \(1.00\) gram mass is converted into energy. Compare this result to the magnitude of the energy changes in chemical reactions that are from about \(-400 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) to \(600 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\).

Glucose is used as fuel in the body according to the reaction $$ \begin{aligned} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+6 \mathrm{O}_{2}(g) \rightarrow 6 \mathrm{CO}_{2}(g) &+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}} &=-2802.5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \end{aligned} $$ During a fever, body temperature rises to about \(39^{\circ} \mathrm{C}\). How many grams of glucose must be burned to raise the temperature of the body from a normal temperature of \(36.8^{\circ} \mathrm{C}\) to \(39.0^{\circ} \mathrm{C}\) for an 82 -kilogram person? Assume that all the energy released as heat from the combustion of glucose is used to heat the body. Assume also that the heat capacity per gram (specific heat) of the body is that of water.

A \(2.50\) -gram sample of powdered zinc is added to \(100.0 \mathrm{~mL}\) of a \(2.0\) -M aqueous solution of hydrochloric acid in a calorimeter with a total heat capacity of \(481 \mathrm{~J} \cdot \mathrm{K}^{-1} .\) The observed increase in temperature is \(12.2 \mathrm{~K}\) at a constant pressure of one bar. Using these data, calculate the value of \(\Delta H_{\mathrm{rxn}}^{\circ}\) for the equation, $$ \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \rightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ which describes the reaction that occurs when the two substances are mixed.

Given that $$ \begin{aligned} \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \rightarrow 2 \mathrm{HF}(g) & \\ \Delta H_{\mathrm{rxn}}^{\circ} &=-546.6 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} \end{aligned} $$ \(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)\) $$ \Delta H_{\mathrm{rxn}}^{\circ}=-571.6 \mathrm{~kJ} \cdot \mathrm{mol}^{-1} $$ calculate the value of \(\Delta H_{\mathrm{rxn}}^{\circ}\) for $$ 2 \mathrm{~F}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow 4 \mathrm{HF}(g)+\mathrm{O}_{2}(g) $$

When \(2.46\) grams of barium reacts with chlorine, \(-15.4 \mathrm{~kJ}\) of energy is evolved as heat. Calculate the energy evolved as heat (in kilojoules) when \(1.00\) mole of barium chloride is formed from barium and chlorine.
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