/*! 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 96 A piece of gold of mass \(0.250 ... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 96

A piece of gold of mass \(0.250 \mathrm{kg}\) and at a temperature of \(75.0^{\circ} \mathrm{C}\) is placed into a \(1.500-\mathrm{kg}\) copper pot containing \(0.500 \mathrm{L}\) of water. The pot and water are at $22.0^{\circ} \mathrm{C}$ before the gold is added. What is the final temperature of the water?

Short Answer

Expert verified
Answer: The final temperature of the water after heat exchange with the gold and the copper pot is approximately \(46.0 ^\circ C\).

Step by step solution

01

Write down the given information

: We are provided with the following information: Mass of gold: \(m_g = 0.250 \, kg\) Initial temperature of gold: \(T_g = 75.0 \, ^\circ C\) Mass of copper pot: \(m_p = 1.50 \, kg\) Initial temperature of pot: \(T_p = 22.0 \, ^\circ C\) Mass of water: \(m_w = 0.5 \, L\), which is equivalent to \(0.5 \, kg\), since water has a density of \(1\, \frac{kg}{L}\) Initial temperature of water: \(T_w = 22.0 \, ^\circ C\) We also need the specific heat capacities of gold, copper, and water: Specific heat of gold: \(c_g = 129 \, \frac{J}{kg \cdot K}\) Specific heat of copper: \(c_p = 385 \, \frac{J}{kg \cdot K}\) Specific heat of water: \(c_w = 4186 \, \frac{J}{kg \cdot K}\)
02

Apply the heat exchange principle

: The total heat gained by the water and the copper pot is equal to the heat lost by the gold. This can be expressed as: \((m_g \cdot c_g \cdot (T_g - T)) + (m_p \cdot c_p \cdot (T_p - T)) = (m_w \cdot c_w \cdot (T - T_w))\)
03

Solve for the final temperature T

: Now we need to solve for T, the final temperature of the water after heat exchange: \((0.250 \cdot 129 \cdot (75.0 - T)) + (1.50 \cdot 385 \cdot (22.0 - T)) = (0.5 \cdot 4186 \cdot (T - 22.0))\) Rearranging the equation and applying the distributive property, we get: \(4833.75 - 32.25T - 866.5T = 2093T - 46183\) Combining like terms, we get: \(-898.75T = -41349.75\) Now, we divide both sides by -898.75 to solve for T: \(T = \frac{-41349.75}{-898.75} \approx 46.0 ^\circ C\)
04

Present the final answer

: The final temperature of the water after the heat exchange with the gold and the copper pot is approximately \(46.0 ^\circ C\).

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Most popular questions from this chapter

A person of surface area \(1.80 \mathrm{m}^{2}\) is lying out in the sunlight to get a tan. If the intensity of the incident sunlight is $7.00 \times 10^{2} \mathrm{W} / \mathrm{m}^{2},$ at what rate must heat be lost by the person in order to maintain a constant body temperature? (Assume the effective area of skin exposed to the Sun is \(42 \%\) of the total surface area, \(57 \%\) of the incident radiation is absorbed, and that internal metabolic processes contribute another \(90 \mathrm{W}\) for an inactive person.)
The power expended by a cheetah is \(160 \mathrm{kW}\) while running at $110 \mathrm{km} / \mathrm{h},$ but its body temperature cannot exceed \(41.0^{\circ} \mathrm{C} .\) If \(70.0 \%\) of the energy expended is dissipated within its body, how far can it run before it overheats? Assume that the initial temperature of the cheetah is \(38.0^{\circ} \mathrm{C},\) its specific heat is \(3.5 \mathrm{kJ} /\left(\mathrm{kg} \cdot^{\circ} \mathrm{C}\right),\) and its mass is \(50.0 \mathrm{kg}\)

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