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A copper calorimeter can with mass 0.100 kg contains 0.160 kg of water and 0.0180 kg of ice in thermal equilibrium at atmospheric pressure. If 0.750 kg of lead at 255\(^\circ\)C is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings.

Short Answer

Expert verified
The final temperature is approximately 20.4°C.

Step by step solution

01

Understand the Problem

We have a copper calorimeter with water and ice initially at 0°C, and we are adding hot lead into it. We need to calculate the final temperature when no heat is lost. This involves heat exchange between the copper can, water, ice, and lead.
02

Define the Specific Heats and Constants

- Specific heat of water (\(c_{water}\) = 4,186 J/kg°C.- Specific heat of ice (\(c_{ice}\) = 2,100 J/kg°C.- Specific heat of copper (\(c_{copper}\) = 390 J/kg°C.- Specific heat of lead (\(c_{lead}\) = 130 J/kg°C.- Latent heat of fusion for ice (\(L_f\) = 334,000 J/kg.
03

Write the Heat Balance Equation

The heat lost by lead equals the heat gained by the copper, water, and ice:\[ m_{lead}c_{lead}(T_{lead} - T_f) = m_{copper}c_{copper}(T_f - T_{initial}) + m_{water}c_{water}(T_f - T_{initial}) + m_{ice}L_f + m_{ice}c_{water}(T_f - 0) \]where \( T_{lead} = 255°C \) and \( T_{initial} = 0°C \).
04

Calculate Heat Changes

- Heat gained by copper: \( Q_{copper} = m_{copper}c_{copper}T_f \)- Heat gained by water (from melted ice): \( Q_{water} = m_{water}c_{water}T_f \)- Heat required to melt ice: \( Q_{fusion} = m_{ice}L_f \)- Heat required to warm melted ice: \( Q_{ice} = m_{ice}c_{water}T_f \)- Heat lost by lead: \( Q_{lead} = m_{lead}c_{lead}(255 - T_f) \).
05

Solve for Final Temperature

By substituting the known values into the heat balance equation:\( 0.75 \times 130 \times (255 - T_f) = 0.10 \times 390 \times T_f + 0.160 \times 4186 \times T_f + 0.018 \times 334000 + 0.018 \times 4186 \times T_f \)Solve this equation for \( T_f \).
06

Final Calculations

Simplify and solve the equation:\( 0.75 \times 130 \times (255 - T_f) = 39T_f + 668.8T_f + 6012 + 75.35T_f \)Combine terms:\( 24375 - 97.5T_f = 1083.15T_f + 6012 \)Calculate \( T_f \) using algebraic methods, you get:\( T_f \approx 20.4°C \).

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Key Concepts

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

Calorimetry
Calorimetry is a science dealing with the measurement of heat exchanges in physical and chemical processes. When objects or substances are mixed or come into contact, they often exchange heat, and calorimetry helps us quantify this exchange. In our exercise, a calorimeter is used, which is a device that measures the heat absorbed or released during these interactions. The important part about calorimetry is that it assumes no heat is lost to the surroundings, which simplifies calculations. To solve problems involving calorimetry, we set up heat balance equations that account for the heat gained and lost by all substances involved. This way, we can determine unknown values such as final temperatures, as seen when adding hot lead to ice and water in the calorimeter.
Heat Exchange
Heat exchange occurs when two or more bodies at different temperatures interact and transfer thermal energy until thermal equilibrium is achieved. Here, the heat lost by the lead is gained by the calorimeter can, the water, and the ice. The principle of conservation of energy indicates that the total amount of heat lost must equal the total amount of heat gained. This concept is crucial for solving the exercise, where the heat from the hot lead helps melt the ice and warm the resulting water as well as the existing water. By understanding and calculating the heat exchanged, we can predict the final state of the system after all exchanges have occurred.
Specific Heat Capacity
Specific heat capacity is a property of a material that tells us how much heat energy is required to change the temperature of one kilogram of the substance by one degree Celsius. Different substances have different specific heat capacities. For example, in our exercise:
  • The specific heat of water is 4,186 J/kg°C.
  • The specific heat of ice is 2,100 J/kg°C.
  • The specific heat of copper is 390 J/kg°C.
  • The specific heat of lead is 130 J/kg°C.
These values are crucial for the calculation of the total heat changes, as they help determine how much energy is required to change the temperatures of the involved substances. The specific heat capacity allows us to set up equations where we can solve for temperatures or energy values when states are changed.
Latent Heat
Latent heat is the heat required to change the state of a substance without changing its temperature. In the context of our problem, the latent heat of fusion is involved because we have ice that starts melting. The latent heat of fusion for ice is 334,000 J/kg, which tells us how much energy is needed to melt the ice into water at 0°C. This latent heat is part of the heat balance equation in the exercise. It is considered a significant component because it accounts for the energy needed to change the ice to water before it can be further warmed. Understanding latent heat helps us acknowledge that not all energy transfer results in a temperature change, as part of it goes into changing the state of matter instead.

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Most popular questions from this chapter

CP A person of mass 70.0 kg is sitting in the bathtub. The bathtub is 190.0 cm by 80.0 cm; before the person got in, the water was 24.0 cm deep. The water is at 37.0\(^\circ\)C. Suppose that the water were to cool down spontaneously to form ice at 0.0\(^\circ\)C, and that all the energy released was used to launch the hapless bather vertically into the air. How high would the bather go? (As you will see in Chapter 20, this event is allowed by energy conservation but is prohibited by the second law of thermodynamics.)

A 4.00-kg silver ingot is taken from a furnace, where its temperature is 750.0\(^\circ\)C, and placed on a large block of ice at 0.0\(^\circ\)C. Assuming that all the heat given up by the silver is used to melt the ice, how much ice is melted?

You put a bottle of soft drink in a refrigerator and leave it until its temperature has dropped 10.0 K. What is its temperature change in (a) F\(^\circ\) and (b) C\(^\circ\)?

You have 750 g of water at 10.0\(^\circ\)C in a large insulated beaker. How much boiling water at 100.0\(^\circ\)C must you add to this beaker so that the final temperature of the mixture will be 75\(^\circ\)C?

A long rod, insulated to prevent heat loss along its sides, is in perfect thermal contact with boiling water (at atmospheric pressure) at one end and with an ice-water mixture at the other (Fig. E17.62). The rod consists of a 1.00-m section of copper (one end in boiling water) joined end to end to a length \(L_2\) of steel (one end in the ice-water mixture). Both sections of the rod have crosssectional areas of 4.00 cm\(^2\). The temperature of the copper- steel junction is 65.0\(^\circ\)C after a steady state has been set up. (a) How much heat per second flows from the boiling water to the ice-water mixture? (b) What is the length \(L_2\) of the steel section?

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