/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 A piece of glass has a temperatu... [FREE SOLUTION] | 91影视

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Chapter 12: Problem 46

A piece of glass has a temperature of \(83.0^{\circ} \mathrm{C} .\) Liquid that has a temperature of \(43.0^{\circ} \mathrm{C}\) is poured over the glass, completely covering it, and the temperature at equilibrium is \(53.0^{\circ} \mathrm{C} .\) The mass of the glass and the liquid is the same. Ignoring the container that holds the glass and liquid and assuming that the heat lost to or gained from the surroundings is negligible, determine the specific heat capacity of the liquid.

Short Answer

Expert verified
The specific heat capacity of the liquid is 2.52 J/g掳C.

Step by step solution

01

Understand the Problem

We are given two objects: a piece of glass and a liquid. Both initially have different temperatures, but after coming into contact, they reach a common equilibrium temperature. Our task is to find the specific heat capacity of the liquid, given that the masses are the same and no heat is lost to the surroundings.
02

Apply the Concept of Heat Transfer

In this closed system, the heat lost by the glass will be equal to the heat gained by the liquid. We can write the equation for heat transfer: \[ m \cdot c_{glass} \cdot (T_{final} - T_{initial, glass}) = m \cdot c_{liquid} \cdot (T_{initial, liquid} - T_{final}) \], where \(m\) is the mass, \(c_{glass}\) and \(c_{liquid}\) are the specific heat capacities, and \(T_{final}\), \(T_{initial, glass}\), \(T_{initial, liquid}\) are the temperatures.
03

Substitute Known Values

Given \(T_{final} = 53.0^{\circ}C\), \(T_{initial, glass} = 83.0^{\circ}C\), \(T_{initial, liquid} = 43.0^{\circ}C\), and the specific heat capacity of glass \(c_{glass} = 0.84\, \text{J/g掳C}\), substitute these into the equation we set up: \[ m \cdot 0.84 \cdot (53.0 - 83.0) = m \cdot c_{liquid} \cdot (43.0 - 53.0) \].
04

Simplify and Solve for Specific Heat Capacity

Since the mass \(m\) cancels out from both sides of the equation, simplify further: \[ 0.84 \cdot (-30.0) = c_{liquid} \cdot (-10.0) \]. Solving for \(c_{liquid}\), we rearrange the equation: \[ c_{liquid} = \frac{0.84 \cdot (-30.0)}{-10.0} = 2.52 \text{ J/g掳C} \].

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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 the process of energy moving from a hotter object to a cooler one. Imagine holding a hot coffee mug; the warmth you feel is due to heat flowing from the mug to your cooler hands. In our exercise, when the liquid is poured over the glass, heat transfers from the warmer glass to the cooler liquid.
This continues until both reach the same temperature. This temperature is called the equilibrium temperature.

In a closed system, like in the exercise where no heat is lost to the surroundings:
  • The heat the glass loses equals the heat the liquid gains.
  • This transfer ensures energy conservation within the system.
  • The equation used in the exercise ensures this balance: \[m \cdot c_{glass} \cdot (T_{final} - T_{initial, glass}) = m \cdot c_{liquid} \cdot (T_{initial, liquid} - T_{final})\]
Equilibrium Temperature
Equilibrium temperature is the common temperature achieved by touching objects after heat exchange. It signifies a balance where heat no longer flows between the two objects as their temperatures are equal. Imagine mixing warm water with cold water; the final temperature falls between their initial temperatures.
In this exercise, both the glass and liquid meet at somewhere in the middle at 53掳C.

  • It reflects a state where energy flow is steady, neither gaining nor losing to the surroundings if isolated.
  • It demonstrates thermal equilibrium, a key principle in thermodynamics.
  • Finding this temperature provides crucial information for calculations, as seen in determining specific heat capacity.
Understanding equilibrium helps frame how energy distributes evenly between interacting elements.
Thermodynamics
Thermodynamics is the branch of physics that studies heat and its relationship to other energy forms. It helps explain how heat moves and reactions operate under energy constraints. In everyday life, understanding how engines work or how refrigerators stay cold connects back to thermodynamic principles like energy conservation and heat transfer.

The exercise illustrates several fundamental thermodynamic principles:
  • Conservation of Energy: Energy is neither created nor destroyed but transferred, illustrating in the equation where heat loss equals heat gain.
  • Closed Systems: With no energy leaving or entering, the system comprises only the glass and liquid, letting us accurately calculate the specific heat capacity of the liquid.
  • Principles Applied: Thermodynamic laws enable us to predict behavior and calculate values like specific heat capacity, useful in designing more efficient processes in engineering and environmental management.
By incorporating thermodynamics, complex interactions in energy conversion and transfer brighten our understanding of natural phenomena.

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Most popular questions from this chapter

When resting, a person has a metabolic rate of about \(3.0 \times 10^{5}\) joules per hour. The person is submerged neck-deep into a tub containing \(1.2 \times 10^{3} \mathrm{kg}\) of water at \(21.00^{\circ} \mathrm{C} .\) If the heat from the person goes only into the water, find the water temperature after half an hour.

The heating element of a water heater in an apartment building has a maximum power output of \(28 \mathrm{kW}\). Four residents of the building take showers at the same time, and each receives heated water at a volume flow rate of \(14 \times 10^{-5} \mathrm{m}^{3} / \mathrm{s}\). If the water going into the heater has a temperature of \(11{ }^{\circ} \mathrm{C},\) what is the maximum possible temperature of the hot water that each showering resident receives?

A constant-volume gas thermometer (see Figures 12.3 and 12.4\()\) has a pressure of \(5.00 \times 10^{3} \mathrm{Pa}\) when the gas temperature is \(0.00^{\circ} \mathrm{C}\). What is the temperature (in \({ }^{\circ} \mathrm{C}\) ) when the pressure is \(2.00 \times 10^{3} \mathrm{Pa} ?\)

One ounce of a well-known breakfast cereal contains 110 Calories (1 food Calorie \(=4186 \mathrm{J}\) ). If \(2.0 \%\) of this energy could be converted by a weight lifter's body into work done in lifting a barbell, what is the heaviest barbell that could be lifted a distance of \(2.1 \mathrm{m} ?\)

Blood can carry excess energy from the interior to the surface of the body, where the energy is dispersed in a number of ways. While a person is exercising, \(0.6 \mathrm{kg}\) of blood flows to the body's surface and releases \(2000 \mathrm{J}\) of energy. The blood arriving at the surface has the temperature of the body's interior, \(37.0^{\circ} \mathrm{C}\). Assuming that blood has the same specific heat capacity as water, determine the temperature of the blood that leaves the surface and returns to the interior.
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