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Stainless steel accessories in cars are usually plated with chromium to give them a shiny surface and to prevent rusting. When \(5.00 \mathrm{~g}\) of chromium at \(23.00^{\circ} \mathrm{C}\) absorbs \(62.5 \mathrm{~J}\) of heat, the temperature increases to \(50.8^{\circ} \mathrm{C}\). What is the specific heat of chromium?

Short Answer

Expert verified
Answer: The specific heat of chromium is approximately 0.449 J/g°C.

Step by step solution

01

List given information

We have the following given information: - Mass of chromium (\(m\)): \(5.00 \mathrm{~g}\) - Initial temperature (\(T_1\)): \(23.00^{\circ} \mathrm{C}\) - Final temperature (\(T_2\)): \(50.8^{\circ} \mathrm{C}\) - Heat absorbed (\(q\)): \(62.5 \mathrm{~J}\)
02

Calculate the change in temperature

Subtract the initial temperature from the final temperature to find the change in temperature. \(\Delta T = T_2 - T_1 = 50.8^{\circ} \mathrm{C} - 23.00^{\circ} \mathrm{C} = 27.8^{\circ} \mathrm{C}\)
03

Write down the formula for heat transfer

We have the formula for heat transfer: \(q = mc\Delta T\) We need to solve for the specific heat, \(c\). Rearrange the formula to make \(c\) the subject: \(c = \frac{q}{m\Delta T}\)
04

Plug in values and compute the specific heat

Now, put the given values and calculated change in temperature into the formula: \(c = \frac{62.5 \mathrm{~J}}{5.00 \mathrm{~g} \times 27.8^{\circ} \mathrm{C}}\) \(c = \frac{62.5}{139} = 0.449 \mathrm{~\frac{J}{g^{\circ} C}}\) So, the specific heat of chromium is approximately \(0.449 \mathrm{~\frac{J}{g^{\circ} C}}\).

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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 an essential concept in physics and engineering, which describes the movement of heat from one body to another. It explains how energy in the form of heat is transferred between objects at different temperatures, achieving a balance, or thermal equilibrium. When considering the specific heat of chromium as in our exercise, we observe heat transfer as the energy that is absorbed by the chromium when its temperature is raised from one value to another.

There are three primary modes of heat transfer: conduction, in which heat moves through a solid material; convection, where heat is carried by moving liquids or gases; and radiation, which involves heat transfer through electromagnetic waves. In our specific case involving chromium, the heat is transferred by conduction as the energy is absorbed into the solid material causing its temperature to rise.

In educational terms, simplifying the concept by relating it to everyday experiences can be helpful. For example, when a metal spoon is placed in hot soup, it will soon become hot due to heat transfer from the soup to the spoon, mainly through conduction. Much like the spoon, the chromium in the exercise absorbs thermal energy, increasing its internal energy and hence its temperature.
Temperature Change
Temperature change refers to the variation in the degree of hotness or coldness measured on a designated scale, like Celsius (°C) in our exercise example. In the context of heat transfer, this change is often a result of energy being added to or removed from a substance. The amount of temperature change experienced by a substance depends on its mass, the amount of heat absorbed or released, and its specific heat capacity.

The specific heat capacity is a property that tells us how much heat energy is required to raise the temperature of one gram of a substance by one degree Celsius. It plays a crucial role in calculating the temperature change and understands the sensitivity of different materials to temperature fluctuations. For instance, if two objects made of different materials absorb the same amount of heat, the one with a lower specific heat capacity will undergo a larger temperature change.

In our chromium example, the substance absorbs a certain amount of heat energy, leading to a measurable temperature increase. By knowing the mass of chromium, the heat absorbed, and the resultant temperature change, we can deduce the specific heat of chromium using the appropriate formulas, providing us with insights into its thermal properties.
Calorimetry
Calorimetry is a branch of thermodynamics dedicated to measuring the amount of heat exchanged in chemical reactions or physical changes. The technique involves using a calorimeter, an instrument designed to isolate and measure these heat exchanges. One common application of calorimetry is determining the specific heat capacity of materials, a property indicating how much energy is needed to raise the temperature of a given mass of a substance by a certain amount.

During the process, the calorimeter could contain a liquid, typically water, in which we immerse an object. The object, initially at a different temperature from the water, will either absorb or lose heat until it reaches thermal equilibrium with the water. The energy change involved is calculated based on the temperature change of the water and the known specific heat capacity of water.

Understanding the fundamentals of calorimetry can enhance comprehension of how specific heat is determined in our chromium example. By comparing the thermal response of chromium to a known reference or through direct measurement, we can gain accurate knowledge about its heat absorbing or releasing properties. Application of calorimetry in practice is often more complex, considering factors like heat losses to the environment, which are specifically accounted for in a proper calorimetric experiment to ensure precise results.

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Most popular questions from this chapter

Isooctane is a primary component of gasoline and gives gasoline its octane rating. Burning \(1.00 \mathrm{~mL}\) of isooctane \((d=0.688 \mathrm{~g} / \mathrm{mL})\) releases \(33.0 \mathrm{~kJ}\) of heat. When \(10.00 \mathrm{~mL}\) of is ooctane is burned in a bomb calorime- ter, the temperature in the bomb rises from \(23.2^{\circ} \mathrm{C}\) to \(66.5^{\circ} \mathrm{C}\). What is the heat capacity of the bomb calorimeter?

Given the following reactions, $$ \begin{aligned} \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{N}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) & \Delta H^{\circ} &=-534.2 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) & \Delta H^{\circ} &=-241.8 \mathrm{~kJ} \end{aligned} $$ calculate the heat of formation of hydrazine.

Microwave ovens convert radiation to energy. A microwave oven uses radiation with a wavelength of \(12.5 \mathrm{~cm}\). Assuming that all the energy from the radiation is converted to heat without loss, how many moles of photons are required to raise the temperature of a cup of water \((350.0 \mathrm{~g}\), specific heat \(=4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) ) from \(23.0^{\circ} \mathrm{C}\) to \(99.0^{\circ} \mathrm{C} ?\)

Equal masses of liquid A, initially at \(100^{\circ} \mathrm{C}\), and liquid B, initially at \(50^{\circ} \mathrm{C}\), are combined in an insulated container. The final temperature of the mixture is \(80^{\circ} \mathrm{C}\). All the heat flow occurs between the two liquids. The two liquids do not react with each other. Is the specific heat of liquid \(A\) larger than, equal to, or smaller than the specific heat of liquid B?

Nitroglycerine, \(\mathrm{C}_{3} \mathrm{H}_{5}\left(\mathrm{NO}_{3}\right)_{3}(l)\), is an explosive most often used in mine or quarry blasting. It is a powerful explosive because four gases \(\left(\mathrm{N}_{2}\right)\) \(\mathrm{O}_{2}, \mathrm{CO}_{2}\), and steam) are formed when nitroglycerine is detonated. In addition, \(6.26 \mathrm{~kJ}\) of heat is given off per gram of nitroglycerine detonated. (a) Write a balanced thermochemical equation for the reaction. (b) What is \(\Delta H\) when \(4.65\) mol of products is formed?

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