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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 chamber with a fixed volume of \(1.0 \mathrm{m}^{3}\) contains a monatomic gas at \(3.00 \times 10^{2} \mathrm{K} .\) The chamber is heated to a temperature of \(4.00 \times 10^{2} \mathrm{K}\). This operation requires \(10.0 \mathrm{J}\) of heat. (Assume all the energy is transferred to the gas.) How many gas molecules are in the chamber?
A cylinder contains 250 L of hydrogen gas \(\left(\mathrm{H}_{2}\right)\) at \(0.0^{\circ} \mathrm{C}\) and a pressure of 10.0 atm. How much energy is required to raise the temperature of this gas to \(25.0^{\circ} \mathrm{C} ?\)
In a physics lab, a student accidentally drops a \(25.0-\mathrm{g}\) brass washer into an open dewar of liquid nitrogen at 77.2 K. How much liquid nitrogen boils away as the washer cools from \(293 \mathrm{K}\) to $77.2 \mathrm{K} ?\( The latent heat of vaporization for nitrogen is \)199.1 \mathrm{kJ} / \mathrm{kg}$.
A blacksmith heats a 0.38 -kg piece of iron to \(498^{\circ} \mathrm{C}\) in his forge. After shaping it into a decorative design, he places it into a bucket of water to cool. If the available water is at \(20.0^{\circ} \mathrm{C},\) what minimum amount of water must be in the bucket to cool the iron to \(23.0^{\circ} \mathrm{C} ?\) The water in the bucket should remain in the liquid phase.
Five ice cubes, each with a mass of \(22.0 \mathrm{g}\) and at a temperature of \(-50.0^{\circ} \mathrm{C},\) are placed in an insulating container. How much heat will it take to change the ice cubes completely into steam?
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