/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 109 A sample of nickel is heated to ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A sample of nickel is heated to \(99.8^{\circ} \mathrm{C}\) and placed in a coffeecup calorimeter containing \(150.0 \mathrm{g}\) water at \(23.5^{\circ} \mathrm{C}\). After the metal cools, the final temperature of metal and water mixture is \(25.0^{\circ} \mathrm{C} .\) If the specific heat capacity of nickel is \(0.444 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\) what mass of nickel was originally heated? Assume no heat loss to the surroundings.

Short Answer

Expert verified
Thus, the mass of nickel that was originally heated is approximately \(28.04 \ \mathrm{g}\).

Step by step solution

01

Determine the heat gained by water

We can calculate the heat gained by water using the following formula: \[Q_{water} = m_{water} \times c_{water} \times \Delta T_{water}\] where: \(Q_{water}\) = Heat gained by water (Joules) \(m_{water}\) = mass of water = 150 g \(c_{water}\) = specific heat capacity of water = 4.18 J/g°C \(\Delta T_{water}\) = change in temperature of water = 25.0 °C - 23.5 °C. Calculate \(\Delta T_{water}\): \[\Delta T_{water} = 25.0 - 23.5 = 1.5^{\circ}\mathrm{C}\] Now, calculate \(Q_{water}\): \[Q_{water} = 150 \times 4.18 \times 1.5 = 940.5\ \mathrm{J}\]
02

Determine the heat lost by nickel

Since there is no heat loss to the surroundings, the heat gained by water should be equal to the heat lost by the nickel. Thus, \[Q_{nickel} = 940.5\ \mathrm{J}\]
03

Calculate the mass of nickel

We can calculate the mass of nickel using the following formula: \[Q_{nickel} = m_{nickel} \times c_{nickel} \times \Delta T_{nickel}\] where: \(Q_{nickel}\) = Heat lost by nickel (Joules) = 940.5 J \(m_{nickel}\) = mass of nickel (grams) \(c_{nickel}\) = specific heat capacity of nickel = 0.444 J/g°C \(\Delta T_{nickel}\) = change in temperature of nickel = 99.8 °C - 25.0 °C. Calculate \(\Delta T_{nickel}\): \[\Delta T_{nickel} = 99.8 - 25.0 = 74.8^{\circ}\mathrm{C}\] Now, calculate \(m_{nickel}\): \[m_{nickel} = \frac{Q_{nickel}}{c_{nickel} \times \Delta T_{nickel}} = \frac{940.5}{0.444 \times 74.8} \approx 28.04 \ \mathrm{g}\] Thus, the mass of nickel that was originally heated is approximately 28.04 grams.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Specific Heat Capacity
Specific heat capacity is a fundamental property of matter that describes how much heat energy is needed to change the temperature of a given mass of a substance by 1 degree Celsius. For example, water has a specific heat capacity of 4.18 J/g°C, which means it requires 4.18 J of heat to raise 1 gram of water by 1°C.

In our exercise, the specific heat capacity informs us how much energy is transferred when the metal and water undergo temperature changes. The specific heat capacity of nickel, as given, is 0.444 J/g°C, signifying it's easier to change the temperature of nickel compared to water.

Understanding specific heat capacity is crucial in calorimetry, a field in thermodynamics that studies heat transfer during physical and chemical processes. The concept helps us compare how different substances respond to added or removed heat.
Heat Transfer
Heat transfer is the process by which thermal energy moves from a hotter object to a cooler one. This movement continues until both objects reach thermal equilibrium, meaning their temperatures become equal.

In the context of the given problem, heat transfer occurs when the hot nickel is placed into cooler water. The nickel cools down, and concurrently, the water warms up until both reach the final equilibrium temperature of 25°C. The calorimetry setup ensures that the heat lost by the nickel equals the heat gained by the water, assuming no heat escapes to the environment.

This scenario illustrates the principle of conservation of energy, where the total energy in the system doesn't change; it merely shifts between objects.
Thermodynamics
Thermodynamics is the branch of physics that deals with the relationships between heat, work, temperature, and energy. It provides us with a framework to understand how energy changes form through different processes.

The exercise on calorimetry is a practical application of the First Law of Thermodynamics, which states that energy cannot be created or destroyed. Instead, it is transferred from one form to another. Here, when nickel cools down, it loses energy in the form of heat, which is then absorbed by the cooler water, leading both to a stable state of thermal equilibrium.

By calculating the mass of nickel based on the heat exchanged, we see the Law of Energy Conservation in action: the heat lost by nickel exactly matches the heat gained by the water. Exercises like these help students understand these principles not only theoretically but also practically, demonstrating how energy transformations follow consistent laws in the universe.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

One way to lose weight is to exercise! Walking briskly at 4.0 miles per hour for an hour consumes about 400 kcal of energy. How many hours would you have to walk at 4.0 miles per hour to lose one pound of body fat? One gram of body fat is equivalent to 7.7 kcal of energy. There are 454 g in 1 lb.

On Easter Sunday, April \(3,1983,\) nitric acid spilled from a tank car near downtown Denver, Colorado. The spill was neutralized with sodium carbonate: $$2 \mathrm{HNO}_{3}(a q)+\mathrm{Na}_{2} \mathrm{CO}_{3}(s) \longrightarrow 2 \mathrm{NaNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)+\mathrm{CO}_{2}(g)$$ a. Calculate \(\Delta H^{\circ}\) for this reaction. Approximately \(2.0 \times\) \(10^{4}\) gal nitric acid was spilled. Assume that the acid was an aqueous solution containing \(70.0 \%\) HNO \(_{3}\) by mass with a density of \(1.42 \mathrm{g} / \mathrm{cm}^{3} .\) What mass of sodium carbonate was required for complete neutralization of the spill, and what quantity of heat was evolved? \((\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{NaNO}_{3}(a q)=-467 \mathrm{kJ} / \mathrm{mol})\) b. According to The Denver Post for April \(4,1983,\) authorities feared that dangerous air pollution might occur during the neutralization. Considering the magnitude of \(\Delta H^{\circ}\) what was their major concern?

A swimming pool, \(10.0 \mathrm{m}\) by \(4.0 \mathrm{m},\) is filled with water to a depth of \(3.0 \mathrm{m}\) at a temperature of \(20.2^{\circ} \mathrm{C}\). How much energy is required to raise the temperature of the water to \(24.6^{\circ} \mathrm{C} ?\)

The specific heat capacity of silver is \(0.24 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\). a. Calculate the energy required to raise the temperature of \(150.0 \mathrm{g}\) Ag from \(273 \mathrm{K}\) to \(298 \mathrm{K}\). b. Calculate the energy required to raise the temperature of 1.0 mole of Ag by \(1.0^{\circ} \mathrm{C}\) (called the molar heat capacity of silver). c. It takes \(1.25 \mathrm{kJ}\) of energy to heat a sample of pure silver from \(12.0^{\circ} \mathrm{C}\) to \(15.2^{\circ} \mathrm{C}\). Calculate the mass of the sample silver.

The enthalpy of combustion of solid carbon to form carbon dioxide is \(-393.7 \mathrm{kJ} / \mathrm{mol}\) carbon, and the enthalpy of combustion of carbon monoxide to form carbon dioxide is \(-283.3 \mathrm{kJ} / \mathrm{mol}\) CO. Use these data to calculate \(\Delta H\) for the reaction $$2 \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}(g)$$

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.