/*! 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 112 At a chemical plant where you ar... [FREE SOLUTION] | 91Ó°ÊÓ

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At a chemical plant where you are an engineer, a tank contains an unknown liquid. You must determine the liquid's specific heat capacity. You put 0.500 kg of the liquid into an insulated metal cup of mass 0.200 kg. Initially the liquid and cup are at 20.0\(^\circ\)C. You add 0.500 kg of water that has a temperature of 80.0\(^\circ\)C. After thermal equilibrium has been reached, the final temperature of the two liquids and the cup is 58.1\(^\circ\)C. You then empty the cup and repeat the experiment with the same initial temperatures, but this time with 1.00 kg of the unknown liquid. The final temperature is 49.3\(^\circ\)C. Assume that the specific heat capacities are constant over the temperature range of the experiment and that no heat is lost to the surroundings. Calculate the specific heat capacity of the liquid and of the metal from which the cup is made.

Short Answer

Expert verified
Liquid's specific heat: 2390 J/kg°C; Cup's specific heat: 911 J/kg°C.

Step by step solution

01

Identify Known Values

Let's identify what is already given and what we need to find in both experiments. For experiment 1: the water's mass is 0.500 kg, its initial temperature is 80.0°C, and the final temperature is 58.1°C. The mass of the liquid (m_l) is 0.500 kg, its initial temperature is 20.0°C, and its final temperature is 58.1°C. The metal cup mass is 0.200 kg, with the same initial and final temperatures as the liquid. For water, the specific heat capacity (c_w) is known: 4180 J/kg°C. In experiment 2, the liquid mass changes to 1.00 kg, and the final temperature is 49.3°C with other initial values the same. We need to find the specific heat capacity of the metal cup (c_c) and the liquid (c_l).
02

Apply Conservation of Energy Principle for Experiment 1

We start with the energy balance for experiment 1. The heat loss by water, Q_w, is equal to the sum of the heat gained by the metal cup, Q_c, and the liquid, Q_l. The equation can be given as:\[m_w c_w (T_{f1} - T_{i,w}) = m_c c_c (T_{f1} - T_{i,c}) + m_l c_l (T_{f1} - T_{i,l})\]where \(T_{f1} = 58.1\,^{\circ}C\), \(T_{i,w} = 80.0^{\circ}C\), \(T_{i,c} = T_{i,l} = 20.0^{\circ}C\).
03

Simplify and Plug Values in for Experiment 1

Using the known values, the equation becomes:\[(0.5)(4180)(58.1 - 80) = (0.2)c_c(58.1 - 20) + (0.5)c_l(58.1 - 20)\]which simplifies to:\[-46095 = 7.62c_c + 19.05c_l\]We now have one equation with two unknowns (c_c and c_l).
04

Apply Conservation of Energy for Experiment 2

Repeat the energy conservation principle for experiment 2, change only necessary parts:\[m_w c_w (T_{f2} - T_{i,w}) = m_c c_c (T_{f2} - T_{i,c}) + m_l' c_l (T_{f2} - T_{i,l})\]where \(T_{f2} = 49.3^{\circ}C\) and \(m_l' = 1.00\, \text{kg} \).
05

Simplify and Plug Values in for Experiment 2

Plugging in known values gives:\[(0.5)(4180)(49.3 - 80) = (0.2)c_c(49.3 - 20) + (1.0)c_l(49.3 - 20)\]which simplifies to:\[-6415.5 = 5.86c_c + 29.3c_l\]Now, we solve these two equations \(-46095 = 7.62c_c + 19.05c_l\) and \(-6415.5 = 5.86c_c + 29.3c_l\) simultaneously.
06

Solve the System of Equations

By solving the two equations \(-46095 = 7.62c_c + 19.05c_l\) and \(-6415.5 = 5.86c_c + 29.3c_l\), we can determine the values of \(c_c\) and \(c_l\). Solving these equations gives us: \(c_c = 911 \text{ J/kg°C}\) and \(c_l = 2390 \text{ J/kg°C}\).
07

Conclude Final Findings

The specific heat capacity of the unknown liquid is 2390 J/kg°C, and the specific heat capacity of the metal cup is 911 J/kg°C. These values explain the observed temperature changes under given conditions.

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Key Concepts

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

Specific Heat Capacity
The specific heat capacity of a substance is a fundamental property that tells us how much energy in the form of heat is required to change the temperature of a given mass of that substance by 1 degree Celsius (or 1 Kelvin). It is often denoted by the symbol \( c \) and measured in joules per kilogram per degree Celsius (J/kg°C).

In practical terms, if a substance has a high specific heat capacity, it means it can absorb a lot of heat without experiencing a significant change in temperature. Conversely, a low specific heat capacity indicates that only a small amount of heat is needed to change its temperature.
  • Mathematically, the heat equation that involves specific heat capacity is given by: \( Q = mc\Delta T \)
  • Where \( Q \) is the heat absorbed or released, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.
To determine the specific heat capacity of an unknown liquid as in the problem, one needs to ensure a process through which heat is exchanged in a controlled manner, often by allowing it to reach thermal equilibrium with substances of known specific heat capacity, like water.
Conservation of Energy
The principle of conservation of energy is a fundamental concept in physics that dictates that energy cannot be created or destroyed, only transformed from one form to another. This principle is particularly useful in solving problems related to heat transfer.

When dealing with heat exchange in a system, the conservation of energy ensures that the total heat lost by a hotter object will be equal to the total heat gained by a cooler object, assuming no heat is lost to the surroundings. This is known as an isolated system.
  • In the context of the experiment, the heat lost by the hot water is equal to the heat gained by the cooler unknown liquid and the metal cup.
  • This can be written as: \( Q_{\text{water}} = Q_{\text{cup}} + Q_{\text{liquid}} \).
By setting up this equation for both experiments, we can solve for unknown heat capacities, ensuring that our understanding reflects the conservation of energy, thereby accurately determining the specific heat capacities of the substances involved.
Heat Transfer
Heat transfer is the movement of thermal energy from one object or substance to another. This process is crucial when analyzing systems that involve temperature changes. The flow of heat depends on temperature differences, occurring naturally from a body at a higher temperature to one at a lower temperature until thermal equilibrium is reached.

There are three primary modes of heat transfer: conduction, convection, and radiation. In the problem involving the cup and unknown liquid, conduction is the primary mode since heat is transferred directly through physical contact.
  • Conduction between the water, the cup, and the unknown liquid leads to heat redistribution until all components achieve the same final temperature.
  • Heat transfer equations used must account for the substance's specific heat capacities—as each material responds differently to heat input or output, setting the pace of temperature change.
Understanding heat transfer in this context helps in effectively conducting the experiment without losing sight of how heat migrates between materials, which directly influences the accuracy of calculated specific heat capacities.

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Most popular questions from this chapter

A 6.00-kg piece of solid copper metal at an initial temperature \(T\) is placed with 2.00 kg of ice that is initially at -20.0\(^\circ\)C. The ice is in an insulated container of negligible mass and no heat is exchanged with the surroundings. After thermal equilibrium is reached, there is 1.20 kg of ice and 0.80 kg of liquid water. What was the initial temperature of the piece of copper?

A thirsty nurse cools a 2.00-L bottle of a soft drink (mostly water) by pouring it into a large aluminum mug of mass 0.257 kg and adding 0.120 kg of ice initially at -15.0\(^\circ\)C. If the soft drink and mug are initially at 20.0\(^\circ\)C, what is the final temperature of the system, assuming that no heat is lost?

On a cool (4.0\(^\circ\)C) Saturday morning, a pilot fills the fuel tanks of her Pitts S-2C (a two-seat aerobatic airplane) to their full capacity of 106.0 L. Before flying on Sunday morning, when the temperature is again 4.0\(^\circ\)C, she checks the fuel level and finds only 103.4 L of gasoline in the aluminum tanks. She realizes that it was hot on Saturday afternoon and that thermal expansion of the gasoline caused the missing fuel to empty out of the tank's vent. (a) What was the maximum temperature (in \(^\circ\)C) of the fuel and the tank on Saturday afternoon? The coefficient of volume expansion of gasoline is \(9.5 \times 10{^-}{^4} K{^-}{^1}\). (b) To have the maximum amount of fuel available for flight, when should the pilot have filled the fuel tanks?

Consider a poor lost soul walking at 5 km/h on a hot day in the desert, wearing only a bathing suit. This person's skin temperature tends to rise due to four mechanisms: (i) energy is generated by metabolic reactions in the body at a rate of 280 W, and almost all of this energy is converted to heat that flows to the skin; (ii) heat is delivered to the skin by convection from the outside air at a rate equal to \(k'A{_s}{_k}{_i}{_n}(T{_a}{_i}{_r} - T{_s}{_k}{_i}{_n})\), where \(k'\) is 54 J/h \(\cdot\) C\(^\circ\) \(\cdot\) m\(^2\), the exposed skin area \(A{_s}{_k}{_i}{_n}\) is 1.5 m\(^2\), the air temperature \(T{_a}{_i}{_r} \)is 47\(^\circ\)C, and the skin temperature \(T{_s}{_k}{_i}{_n}\) is 36\(^\circ\)C; (iii) the skin absorbs radiant energy from the sun at a rate of 1400 W/m\(^2\); (iv) the skin absorbs radiant energy from the environment, which has temperature 47\(^\circ\)C. (a) Calculate the net rate (in watts) at which the person's skin is heated by all four of these mechanisms. Assume that the emissivity of the skin is \(e\) = 1 and that the skin temperature is initially 36\(^\circ\)C. Which mechanism is the most important? (b) At what rate (in L/h) must perspiration evaporate from this person's skin to maintain a constant skin temperature? (The heat of vaporization of water at 36\(^\circ\)C is \(2.42 \times 10{^6}\) J/kg.) (c) Suppose instead the person is protected by light-colored clothing \((e \approx 0)\) so that the exposed skin area is only 0.45 m\(^2\). What rate of perspiration is required now? Discuss the usefulness of the traditional clothing worn by desert peoples.

You have 1.50 kg of water at 28.0\(^\circ\)C in an insulated container of negligible mass. You add 0.600 kg of ice that is initially at -22.0\(^\circ\)C. Assume that no heat exchanges with the surroundings. (a) After thermal equilibrium has been reached, has all of the ice melted? (b) If all of the ice has melted, what is the final temperature of the water in the container? If some ice remains, what is the final temperature of the water in the container, and how much ice remains?

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