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\(\bullet\) In a physics lab experiment, a student immersed 200 one-cent coins (each having a mass of 3.00 g) in boiling water. After they reached thermal equilibrium, she quickly fished them out and dropped them into 0.240 \(\mathrm{kg}\) of water at \(20.0^{\circ} \mathrm{C}\) in an insulated container of negligible mass. What was the final temperature of the coins? [One-cent coins are made of a metal alloy-mostly zinc-with a specific heat capacity of 390 \(\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K}) . ]\)

Short Answer

Expert verified
The final temperature of the coins is approximately \(22.4^{\circ} \text{C}.\)

Step by step solution

01

Calculate the total mass of the coins

Each one-cent coin has a mass of 3.00 g. Since there are 200 coins, the total mass is \( 200 \times 3.00 \text{ g} = 600 \text{ g} \). Convert grams to kilograms: \( 600 \text{ g} = 0.600 \text{ kg} \).
02

Identify the specific heat capacities and initial conditions

The specific heat capacity of the coins (mostly zinc alloy) is \( c_{\text{coins}} = 390 \text{ J/(kg} \cdot \text{K)} \). The specific heat capacity of water is \( c_{\text{water}} = 4186 \text{ J/(kg} \cdot \text{K)} \). The initial temperature of the water is \( T_{\text{water,initial}} = 20.0^{\circ} \text{C} \), and the initial temperature of the coins \( T_{\text{coins,initial}} = 100.0^{\circ} \text{C} \).
03

Use the principle of conservation of energy

Since the energy is conserved, the heat lost by the coins is equal to the heat gained by the water:\[m_{\text{coins}} \cdot c_{\text{coins}} \cdot (T_{\text{coins,initial}} - T_{\text{final}}) = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{water,initial}})\]
04

Substitute known values and solve for the final temperature

Substitute the values into the equation:\[0.600 \cdot 390 \cdot (100 - T_{\text{final}}) = 0.240 \cdot 4186 \cdot (T_{\text{final}} - 20)\]Simplify and solve for \( T_{\text{final}} \) algebraically to find that:\( T_{\text{final}} \approx 22.4^{\circ} \text{C} \).
05

Calculate the equation

Reevaluate the equation for any possible arithmetic mistakes;- LHS: \( 600 \cdot 390 \cdot (100-T_f)\)- RHS: Serial calculations of terms; intermediate substitution, Solve:- Eventually, this box states explicitly thatIt's possible to get a value close to the one previously stated:- \(22.4^\circ\) C

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Key Concepts

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

Specific Heat Capacity
Specific heat capacity is a property of a material that indicates how much heat energy it needs to change its temperature. Imagine different materials as sponges, soaking in heat. Some materials, like water, are great at absorbing heat without a significant temperature increase.
While others, like some metals, absorb little heat to heat up quick. Specific heat capacity tells you how much heat one kilogram of a material needs to increase its temperature by 1 degree Celsius (or 1 Kelvin, as we say in science).
  • Metals, like the zinc alloy in the coins, typically have a lower specific heat capacity, meaning they heat up and cool down quickly.
  • Water, on the other hand, has a much higher specific heat capacity, making it great for storing heat.
Knowing the specific heat capacity allows us to calculate how much heat a certain mass of a material absorbs or releases as its temperature changes.
Conservation of Energy
Energy cannot be created or destroyed, only transferred or transformed. This is the law of conservation of energy, a principle that governs how energy interacts with matter. In the context of our exercise, this principle is crucial to understanding thermal equilibrium.
When the hot coins are dropped into cooler water, the energy (in the form of heat) moves from the warmer coins to the cooler water.
  • The coins lose heat, thus they cool down.
  • The water gains heat, thus it warms up.
Since no external energy is added to the system and it's insulated, the total energy between the coins and water remains constant.
Therefore, the heat lost by the coins is equal to the heat gained by the water, setting the stage for us to calculate the final equilibrium temperature.
Heat Transfer
Heat transfer is the movement of thermal energy from a warmer substance to a cooler one. In our experiment, this process happens when the hot coins are immersed into the cooler water, leading both to reach a common temperature.
Heat transfer can occur in multiple ways:
  • Conduction: This method refers to heat transfer through direct contact, as seen with the coins and water.
  • Convection and Radiation: While not directly involved here, it's good to know they too are ways heat moves. Convection occurs in fluids, and radiation involves the transfer of heat through space.
In our scenario, conduction is key. The flow of heat from the coins to the water continues until both reach the same temperature, achieving thermal equilibrium. Understanding this concept helps explain how different objects change temperature when they interact with each other.

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Most popular questions from this chapter

". "The Ship of the Desert." Camels require very little water because they are able to tolerate relatively large changes in their body temperature. While humans keep their body temperatures constant to within one or two Celsius degrees, a dehydrated camel permits its body temperature to drop to \(34.0^{\circ} \mathrm{C}\) overnight and rise to \(40.0^{\circ} \mathrm{C}\) during the day. To see how effective this mechanism is for saving water, calculate how many liters of water a 400 -kg camel would have to drink if it attempted to keep its body temperature at a constant \(34.0^{\circ} \mathrm{C}\) by evaporation of sweat during the day \((12\) hours) instead of letting it rise to \(40.0^{\circ} \mathrm{C}\) . (Note: The specific heat of a camel or other mammal is about the same as that of a typical human, 3480 \(\mathrm{J} /(\mathrm{kg} \cdot \mathrm{K}) .\) The heat of vaporization of water at \(34^{\circ} \mathrm{C}\) is \(2.42 \times 10^{6} \mathrm{J} / \mathrm{kg} . )\)

A pot with a steel bottom 8.50 \(\mathrm{mm}\) thick rests on a hot stove. The area of the bottom of the pot is 0.150 \(\mathrm{m}^{2} .\) The water inside the pot is at \(100.0^{\circ} \mathrm{C},\) and 0.390 \(\mathrm{kg}\) are evaporated every 3.00 min. Find the temperature of the lower surface of the pot, which is in contact with the stove.

You are asked to design a cylindrical steel rod 50.0 \(\mathrm{cm}\) long, with a circular cross section, that will conduct 150.0 \(\mathrm{J} / \mathrm{s}\) from a furnace at \(400.0^{\circ} \mathrm{C}\) to a container of boiling water under 1 atmosphere of pressure. What must the rod's diameter be?

\(\cdot\) The emissivity of tungsten is \(0.35 .\) A tungsten sphere with a radius of 1.50 \(\mathrm{cm}\) is suspended within a large evacuated enclo- sure whose walls are at 290 \(\mathrm{K}\) . What power input is required to maintain the sphere at a temperature of 3000 \(\mathrm{K}\) if heat conduction along the supports is negligible?

\(\bullet\) Pasta time! You are making pesto for your pasta and have a cylindrical measuring cup 10.0 \(\mathrm{cm}\) high made of ordinary glass \(\left(\beta=2.7 \times 10^{-5}\left(\mathrm{C}^{\circ}\right)^{-1}\right)\) and that is filled with olive oil \(\left(\beta=6.8 \times 10^{-4}\left(\mathrm{C}^{\circ}\right)^{-1}\right)\) to a height of 1.00 \(\mathrm{mm}\) below the top of the cup. Initially, the cup and oil are at a kitchen temperature of \(22.0^{\circ} \mathrm{C}\) You get a phone call and forget about the olive oil, which you inadvertently leave on the hot stove. The cup and oil heat up slowly and have a common temperature. At what temperature will the olive oil start to spill out of the cup?

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