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Chapter 6: Problem 42

One piece of copper jewelry at \(105^{\circ} \mathrm{C}\) has twice the mass of another piece at \(45^{\circ} \mathrm{C}\). Both are placed in a calorimeter of negligible heat capacity. What is the final temperature inside the calorimeter \((c\) of copper \(=0.387 \mathrm{~J} / \mathrm{g} \cdot \mathrm{K}) ?\)

Short Answer

Expert verified
The final temperature is 85掳C.

Step by step solution

01

Identify Variables

Let the masses of two copper pieces be 2m (hot piece) and m (cold piece). Let the final temperature be T.
02

Write Heat Lost Equation

Heat loss by the hot piece is calculated by \[Q_{\text{loss}} = 2m \times c \times (105 - T).\]
03

Write Heat Gained Equation

Heat gain by the cold piece is calculated by \[Q_{\text{gain}} = m \times c \times (T - 45).\]
04

Set Up Heat Exchange Equation

Since heat lost by the hot piece equals heat gained by the cold piece, \[2m \times c \times (105 - T) = m \times c \times (T - 45).\]
05

Simplify the Equation

Cancel out the common terms (m and c), and solve for T:\[2 \times (105 - T) = T - 45,\]\[210 - 2T = T - 45,\]\[210 + 45 = 3T,\]\[255 = 3T,\]\[T = 85.\]
06

Final Answer

The final temperature inside the calorimeter is 85掳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 measure of how much heat energy is required to raise the temperature of a substance. It is usually denoted by the symbol \(c\). The unit of specific heat capacity is Joules per gram per degree Kelvin (\( \text{J/g} \bullet \text{K} \)). A larger specific heat capacity means a substance can absorb more heat without changing temperature significantly.
In our exercise, the specific heat capacity of copper is given as \(0.387 \text{~J/g} \bullet \text{K}\). To label it clearly:
  • For copper, \(c = 0.387 \text{~J/g} \bullet \text{K}\).
heat transfer
Heat transfer is the process by which thermal energy moves from a hotter object to a cooler one. Heat always flows from regions of higher temperature to regions of lower temperature until thermal equilibrium is reached.
There are three primary methods of heat transfer:
  • Conduction: Heat transfer through a solid material.
  • Convection: Heat transfer through a fluid (liquid or gas).
  • Radiation: Heat transfer through electromagnetic waves.

In our problem, we focus on conduction, where heat energy is exchanged between two pieces of copper. We calculate heat lost and heat gained by each piece using the formulas:
\[Q_{\text{loss}} = 2m \times c \times (105 - T)\]
\[Q_{\text{gain}} = m \times c \times (T - 45)\]
thermal equilibrium
Thermal equilibrium is the condition where all involved temperatures become equal, and there is no net heat transfer between objects. In our problem, both copper pieces in the calorimeter ultimately reach the same final temperature.
The concept is crucial as it allows us to set up the heat exchange equation:
\[2m \times c \times (105 - T) = m \times c \times (T - 45)\]
By solving this equation, we find the final temperature inside the calorimeter when the two pieces of copper have the same temperature:
  • The final temperature calculated is \(85^{\text{o}}\text{C}\).

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