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Chapter 12: Problem 48

Two bars of identical mass are at \(25^{\circ} \mathrm{C}\). One is made from glass and the other from another substance. The specific heat capacity of glass is \(840 \mathrm{J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right) .\) When identical amounts of heat are supplied to each, the glass bar reaches a temperature of \(88^{\circ} \mathrm{C},\) while the other bar reaches \(250.0^{\circ} \mathrm{C}\). What is the specific heat capacity of the other substance?

Short Answer

Expert verified
The specific heat capacity of the unknown substance is 235.2 J/kg掳C.

Step by step solution

01

Understand the problem

We have two bars of identical mass and specific heat capacities. We need to find the specific heat capacity of the unknown substance after heating both materials from an initial temperature of 25掳C. We know the final temperature for glass and the unknown substance, and we're given the specific heat capacity for glass.
02

Use the formula for heat transfer

The formula to calculate the heat absorbed is \( Q = mc\Delta T \), where \( Q \) is heat absorbed, \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the temperature change. We'll apply it to both materials to find the unknown specific heat capacity.
03

Calculate the temperature change for each material

For the glass bar, \( \Delta T = 88^{\circ} C - 25^{\circ} C = 63^{\circ} C \). For the other substance, \( \Delta T = 250^{\circ} C - 25^{\circ} C = 225^{\circ} C \).
04

Expression of heat supplied to each material

Assume heat supplied to both substances is \(Q\) (same for both). For glass: \( Q = m \times 840 \, \mathrm{J/kg^{\circ}C} \times 63^{\circ} C \). For other substance: \( Q = m \times c_{\text{unknown}} \times 225^{\circ} C \).
05

Solve equations to find specific heat of unknown substance

Equating both expressions for \( Q \) as they are identical, we have:\[ m \times 840 \times 63 = m \times c_{\text{unknown}} \times 225 \].The mass \( m \) cancels out, simplifying to:\[ 840 \times 63 = c_{\text{unknown}} \times 225 \].Solving for \( c_{\text{unknown}} \), we find:\[ c_{\text{unknown}} = \frac{840 \times 63}{225} = 235.2 \, \mathrm{J/kg^{\circ}C} \].

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Key Concepts

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

Heat Transfer
Heat transfer is all about the movement of energy from one body or material to another. It's the reason why a cold spoon gets warm when placed in a hot cup of soup. In the context of physics, heat transfer is quantified through the concept of heat (often represented by the letter \(Q\)).
Here, heat acts as the energy that flows because of the temperature difference. In our exercise, identical amounts of heat are supplied to two bars 鈥� one made of glass and the other one made of an unknown substance.
Understanding how heat moves and how it affects the objects involved is the key to grasping many real-world and scientific phenomena, from cooking to meteorology.
Temperature Change
When heat is supplied to a substance, it often results in a change in temperature. This change, noted as \(\Delta T\), is calculated by subtracting the initial temperature from the final temperature.
For instance, the glass bar in the exercise started at \(25^{\circ} C\) and ended at \(88^{\circ} C\), leading to a temperature change of \(63^{\circ} C\).
Recognizing the temperature change is crucial as it directly impacts how much heat is absorbed by the object, contingent on its specific heat capacity.
  • Initial temperature (glass): \(25^{\circ} C\)
  • Final temperature (glass): \(88^{\circ} C\)
  • Temperature change (glass): \(63^{\circ} C\)
  • Initial temperature (unknown substance): \(25^{\circ} C\)
  • Final temperature (unknown substance): \(250^{\circ} C\)
  • Temperature change (unknown substance): \(225^{\circ} C\)
Heat Absorbed
The amount of heat absorbed by a substance is vital for understanding how temperatures change. To calculate the heat absorbed, the formula \( Q = mc\Delta T \) is used, where \(Q\) represents the heat absorbed, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the temperature change.
Each material absorbs heat differently based on its specific heat capacity. Given that both bars are heated with an identical amount of heat, the temperature response varies because of their differing specific heat capacities. This equation allows us to pinpoint how much energy causes specific temperature shifts in different materials.
Identical Mass
In problems involving heat transfer and temperature change, the assumption of identical mass simplifies calculations. Here, both bars are stated to have identical mass.
This is crucial, as identical mass ensures the only variables influencing heat transfer are the specific heat capacities and temperature changes. This makes the mass value cancel out when equating heat absorbed by the bars, making our calculation straightforward.
Such simplifications are often used in physics problems to focus on learning the primary concepts without unnecessary complications.

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Most popular questions from this chapter

You and your team are testing a device that is to be submerged in the cold waters of Antarctica. It is designed to rotate a small wheel at a very precise rate. One component of the device consists of a steel wheel (diameter \(D_{\text {sieel }}=3.0000 \mathrm{cm}\) at \(\left.T=78.400^{\circ} \mathrm{F}\right)\) and a large aluminum wheel that drives it (diameter \(D_{\mathrm{A}}=150.000 \mathrm{cm}\) at \(T=78.400^{\circ} \mathrm{F}\) ). (a) If the aluminum wheel rotates at 35.0000 RPM, at what rate does the smaller, steel wheel turn while both wheels are at \(T=78.400^{\circ} \mathrm{F} ?\) (b) You submerge the device in a large vat of water held at \(32.800^{\circ} \mathrm{C},\) simulating its working environment in the Antarctic seas. Assuming the larger, aluminum driving wheel still rotates at \(35.0000 \mathrm{RPM},\) at what rate does the smaller, steel wheel rotate when the device is submerged in the cold water? (c) How should you adjust the rate of the aluminum driving wheel so that the steel wheel rotates at the same rate as it had at \(T=78.400^{\circ} \mathrm{F} ?\) Note: \(\alpha_{\mathrm{Al}}=23 \times 10^{-6}\) and \(\alpha_{\text {steel }}=12 \times 10^{-6}\)

Water at \(23.0^{\circ} \mathrm{C}\) is sprayed onto \(0.180 \mathrm{kg}\) of molten gold at \(1063^{\circ} \mathrm{C}\) (its melting point). The water boils away, forming steam at \(100.0^{\circ} \mathrm{C}\) and leaving solid gold at \(1063^{\circ} \mathrm{C} .\) What is the minimum mass of water that must be used?

A copper-constantan thermocouple generates a voltage of \(4.75 \times\) \(10^{-3}\) volts when the temperature of the hot junction is \(110.0^{\circ} \mathrm{C}\) and the reference junction is kept at a temperature of \(0.0^{\circ} \mathrm{C} .\) If the voltage is proportional to the difference in temperature between the junctions, what is the temperature of the hot junction when the voltage is \(1.90 \times 10^{-3}\) volts?

A 42 -kg block of ice at \(0^{\circ} \mathrm{C}\) is sliding on a horizontal surface. The initial speed of the ice is \(7.3 \mathrm{m} / \mathrm{s}\) and the final speed is \(3.5 \mathrm{m} / \mathrm{s} .\) Assume that the part of the block that melts has a very small mass and that all the heat generated by kinetic friction goes into the block of ice. Determine the mass of ice that melts into water at \(0^{\circ} \mathrm{C}\).

During an all-night cram session, a student heats up a one-half liter \(\left(0.50 \times 10^{-3} \mathrm{m}^{3}\right)\) glass (Pyrex) beaker of cold coffee. Initially, the temperature is \(18^{\circ} \mathrm{C},\) and the beaker is filled to the brim. A short time later when the student returns, the temperature has risen to \(92^{\circ} \mathrm{C} .\) The coefficient of volume expansion of coffee is the same as that of water. How much coffee (in cubic meters) has spilled out of the beaker?
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