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A piece of chromium metal with a mass of \(24.26 \mathrm{g}\) is heated in boiling water to \(98.3^{\circ} \mathrm{C}\) and then dropped into a coffee-cup calorimeter containing 82.3 g of water at \(23.3^{\circ}\) C. When thermal equilibrium is reached, the final temperature is \(25.6^{\circ} \mathrm{C} .\) Calculate the specific heat capacity of chromium.

Short Answer

Expert verified
The specific heat capacity of chromium is approximately \(0.448 \text{ J/g}^\circ\text{C}\).

Step by step solution

01

Understanding Heat Transfer

When the heated chromium is dropped into the water, heat transfer occurs from the chromium to the water until they reach thermal equilibrium. The heat lost by chromium is equal to the heat gained by the water.
02

Write the Heat Transfer Equations

The heat gained or lost is calculated using the formula: \[ q = m \cdot c \cdot \Delta T \]where \( q \) is the heat (in joules), \( m \) is the mass (in grams), \( c \) is the specific heat capacity (in \(\text{J/g}^\circ\text{C}\)), and \( \Delta T \) is the change in temperature (in degrees Celsius).
03

Calculate Heat Gained by Water

The mass of water is \(82.3 \text{ g}\), the specific heat capacity of water is \(4.18 \text{ J/g}^\circ\text{C}\), and its temperature change is from \(23.3^{\circ}\text{C}\) to \(25.6^{\circ}\text{C}\).\[ q_\text{water} = 82.3 \cdot 4.18 \cdot (25.6 - 23.3) \]\[ q_\text{water} = 82.3 \cdot 4.18 \cdot 2.3 \]\[ q_\text{water} = 790.4034 \text{ J} \]
04

Calculate Heat Lost by Chromium

The heat lost by chromium \( q_\text{chromium} \) is equal in magnitude to the heat gained by water but with an opposite sign, i.e., \( q_\text{chromium} = -790.4034 \text{ J} \). We need to find the specific heat capacity \( c_\text{chromium} \). The mass of chromium is \(24.26 \text{ g}\), and its temperature change is from \(98.3^{\circ}\text{C}\) to \(25.6^{\circ}\text{C}\).\[ \Delta T_\text{chromium} = 25.6 - 98.3 = -72.7^{\circ}\text{C} \].
05

Solve for Specific Heat Capacity of Chromium

Using the formula \( q = m \cdot c \cdot \Delta T \) and the known values:\[ -790.4034 = 24.26 \cdot c_\text{chromium} \cdot (-72.7) \]Solving for \( c_\text{chromium} \):\[ c_\text{chromium} = \frac{790.4034}{24.26 \times 72.7} \]\[ c_\text{chromium} \approx \frac{790.4034}{1763.202} \approx 0.448 \text{ J/g}^\circ\text{C} \]
06

Interpretation of Results

The calculated specific heat capacity of chromium is approximately \(0.448 \text{ J/g}^\circ\text{C}\), which implies how much energy is required to change the temperature of one gram of chromium by one degree Celsius.

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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 the process of thermal energy moving from one object to another due to a difference in temperature. In our exercise, when the hot piece of chromium is placed into cooler water, heat energy is transferred from the chromium to the water. This flow of heat continues until both the metal and water reach the same temperature, achieving a state of balance. The principle of conservation of energy underlies heat transfer, meaning the total energy lost by the chromium is exactly equal to the energy gained by the water. This principle helps us write a crucial equation for solutions in calorimetry problems.
Thermal Equilibrium
Thermal equilibrium is reached when two objects in contact with each other exchange no more heat. They have, at this point, the same temperature. In calorimetry experiments like this one, achieving thermal equilibrium allows us to determine the heat transfer involved. Knowing the final equilibrium temperature, you can calculate the amount of heat exchanged. For example, in this exercise, both chromium and water settle at a final temperature of 25.6°C. It indicates that they have reached thermal equilibrium.
Calorimetry
Calorimetry is the science of measuring heat transfer between substances. It uses a device called a calorimeter to ensure that no heat is lost to the surroundings. In this scenario, a coffee-cup calorimeter was utilized. Calorimetry is based on the idea that the heat gained by one substance must equal the heat lost by another. Thus, by knowing the heat capacity of water and its temperature change, it is possible to compute the heat transferred. With this step done, you can use the heat exchange to find the unknown specific heat capacity of another substance—in this exercise, the chromium sample.
Temperature Change
Temperature change (\(\Delta T\)) is a critical factor in computing heat transfer. It represents how much an object's temperature shifts as it gains or loses heat. You calculate it by subtracting the initial temperature from the final temperature. In our problem, for chromium, the temperature drops from 98.3°C to 25.6°C. The change is reflected as negative 72.7°C, indicating a loss of heat. Each substance's heat capacity dictates how much its temperature shifts based on this transferred energy. Calculating this change accurately helps you find the specific heat capacities used in energy balance equations.

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Most popular questions from this chapter

The internal energy of a gas decreases by \(1.65 \mathrm{kJ}\) when it transfers \(1.87 \mathrm{kJ}\) of energy in the form of heat to the surroundings. (a) Calculate the work done by the gas on the surroundings. (b) Does the volume of gas increase or decrease?

A 25.0 -mL. sample of benzene at \(19.9^{\circ} \mathrm{C}\) was cooled to its melting point, \(5.5^{\circ} \mathrm{C},\) and then frozen. How much energy was given off as heat in this process? (The density of benzene is \(0.80 \mathrm{g} / \mathrm{mL},\) its specific heat capacity is \(1.74 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\) and its heat of fusion is \(127 \mathrm{J} / \mathrm{g} .\) )

A volume of 1.50 L of argon gas is confined in a cylinder with a movable piston under a constant pressure of \(1.22 \times 10^{5}\) Pa. When \(1.25 \mathrm{kJ}\) of energy in the form of heat is transferred from the surroundings to the gas, the internal energy of the gas increases by 1.11 kJ. What is the final volume of argon gas in the cylinder?

Ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH},\) boils at \(78.29^{\circ} \mathrm{C} .\) How much energy, in joules, is required to raise the temperature of \(1.00 \mathrm{kg}\) of ethanol from \(20.0^{\circ} \mathrm{C}\) to the boiling point and then to change the liquid to vapor at that temperature? (The specific heat capacity of liquid ethanol is \(2.44 \mathrm{J} / \mathrm{g} \cdot \mathrm{K},\) and its enthalpy of vaporization is \(855 \mathrm{J} / \mathrm{g} .\) )

Suppose that an inch \((2.54 \mathrm{cm})\) of rain falls over a square mile of ground \(\left(2.59 \times 10^{6} \mathrm{m}^{2}\right) .\) (Density of water is \(1.0 \mathrm{g} / \mathrm{cm}^{3} .\) ) The enthalpy of vaporization of water at \(25^{\circ} \mathrm{C}\) is \(44.0 \mathrm{kJ} / \mathrm{mol} .\) How much energy is transferred as heat to the surroundings from the condensation of water vapor in forming this quantity of liquid water? (The huge number tells you how much energy is "stored" in water vapor and why we think of storms as such great forces of energy in nature. It is interesting to compare this result with the energy given off, \(4.2 \times 10^{6} \mathrm{kJ},\) when a ton of dynamite explodes.

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