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Titanium is a metal used in jet engines. Its specific heat is \(0.523 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). If \(5.88 \mathrm{~g}\) of titanium absorbs \(4.78 \mathrm{~J}\), what is the change in temperature?

Short Answer

Expert verified
Answer: The change in temperature of the titanium is approximately 1.47°C.

Step by step solution

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01

Write the given values.

We are given: Specific heat of titanium (c) = 0.523 J/g °C Mass of titanium (m) = 5.88 g Energy absorbed (q) = 4.78 J
02

Write the formula for calculating the change in temperature (ΔT).

The formula for calculating the change in temperature is: \(q = mcΔT\)
03

Rearrange the formula for ΔT.

We need to find ΔT, so we will rearrange the formula: \(ΔT = \frac{q}{mc}\)
04

Plug in the given values and calculate ΔT.

Now we will substitute the given values for q, m, and c in the formula: \(ΔT = \frac{4.78 \mathrm{~J}}{5.88 \mathrm{~g} \times 0.523 \mathrm{~J / g\cdot ^{\circ}C}}\)
05

Solve for ΔT.

Now, calculate the value of ΔT: \(ΔT = \frac{4.78 \mathrm{~J}}{ (5.88 \mathrm{~g}) \times (0.523 \mathrm{~J / g\cdot ^{\circ}C})} \approx 1.47^{\circ} C\) The change in temperature of the titanium is approximately 1.47°C.

Key Concepts

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

Thermal Energy
Thermal energy refers to the total energy present within a substance due to the random motion of its molecules. It is important to distinguish between thermal energy and temperature, as they are not the same. While temperature measures the average kinetic energy of particles in a substance, thermal energy accounts for both the temperature and the number of particles, or the mass, in the substance.
In our exercise, thermal energy is involved when the titanium absorbs the heat, which then causes a change in its temperature. This energy is measured in joules (J), which is a standard unit of energy in physics and other sciences. When titanium absorbs 4.78 J of energy in this problem, we are witnessing the process of thermal energy being transferred to the metal thereby falling under the principle of thermal dynamics.
Understanding thermal energy allows us to comprehend how energy interchange takes place in various materials, and why the heat capacity of substances, like titanium in this case, is crucial for determining how much temperature change occurs when heat is absorbed.
Temperature Change
Temperature change (\(\Delta T\)) is the measure of how much the temperature of an object increases or decreases after absorbing or releasing heat. In this exercise, we aim to calculate the temperature change of titanium upon the absorption of a certain amount of energy.
To calculate the temperature change, we use the formula:\[\Delta T = \frac{q}{mc}\]This equation states that the change in temperature depends on the amount of energy transferred (\(q\)), the mass of the substance (\(m\)), and the specific heat capacity of the substance (\(c\)).
In the given task, the titanium absorbs 4.78 J of energy, and we need to find how much this affects the temperature. The available data: mass (5.88 g) and specific heat (0.523 J/g°C), need to be substituted into the formula to solve for the temperature change. Following this calculation, it becomes evident that the 4.78 J increases the temperature by approximately 1.47°C.
Heat Capacity
Heat capacity is the total amount of heat energy required to change the temperature of an object by a given amount, typically one degree Celsius. It is intimately related to the specific heat, which describes the heat capacity per unit mass of a substance. The specific heat capacity provides insight into how resistant a material is to changing temperature.
For titanium, its specific heat capacity is given as 0.523 J/g°C in the exercise. This tells us that it takes 0.523 J of energy to raise 1 gram of titanium by 1°C. This specific heat capacity value, along with the mass of the titanium, plays a crucial role in our calculations because it influences how much energy is needed to achieve a certain temperature change.
  • Higher specific heat means the material will require more energy to increase its temperature.
  • Lower specific heat means it will heat up with less energy input.
In practical terms, understanding heat capacity helps in designing systems that manage heat exchange efficiently, like those found in engines and other mechanical applications where metals like titanium are used.

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Most popular questions from this chapter

Gold has a specific heat of \(0.129 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\). When a \(5.00-\mathrm{g}\) piece of gold absorbs \(1.33\) J of heat, what is the change in temperature?

When \(35.0 \mathrm{~mL}\) of \(1.43 \mathrm{M} \mathrm{NaOH}\) at \(22.0^{\circ} \mathrm{C}\) is neutralized by \(35.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) also at \(22.0^{\circ} \mathrm{C}\) in a coffee-cup calorimeter, the temperature of the final solution rises to \(31.29^{\circ} \mathrm{C}\). Assume that the specific heat of all solutions is \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\), that the density of all solutions is \(1.00 \mathrm{~g} / \mathrm{mL}\), and that volumes are additive. (a) Calculate \(q\) for the reaction. (b) Calculate \(q\) for the neutralization of one mole of \(\mathrm{NaOH}\).

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?

Urea, \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\), is used in the manufacture of resins and glues. When \(5.00 \mathrm{~g}\) of urea is dissolved in \(250.0 \mathrm{~mL}\) of water \((d=1.00 \mathrm{~g} / \mathrm{mL})\) at \(30.0^{\circ} \mathrm{C}\) in a coffee-cup calorimeter, \(27.6 \mathrm{~kJ}\) of heat is absorbed. (a) Is the solution process exothermic? (b) What is \(q_{\mathrm{H}_{2} \mathrm{O}}\) ? (c) What is the final temperature of the solution? (Specific heat of water is \(4.18 \mathrm{~J} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\).) (d) What are the initial and final temperatures in \({ }^{\circ} \mathrm{F}\) ?

Draw a cylinder with a movable piston containing six molecules of a liquid. A pressure of 1 atm is exerted on the piston. Next draw the same cylinder after the liquid has been vaporized. A pressure of one atmosphere is still exerted on the piston. Is work done on the system or by the system?

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