/*! 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 58 A glass vial containing a 16.0 -... [FREE SOLUTION] | 91Ó°ÊÓ

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A glass vial containing a 16.0 -g sample of an enzyme is cooled in an ice bath. The bath contains water and 0.120 \(\mathrm{kg}\) of ice. The sample has specific heat 2250 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) ; the glass vial has mass 6.00 \(\mathrm{g}\) and specific heat 2800 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K} .\) How much ice melts in cooling the enzyme sample from room temperature \(\left(19.5^{\circ} \mathrm{C}\right)\) to the temperature of the ice bath?

Short Answer

Expert verified
Approximately 3.08 g of ice melts.

Step by step solution

01

Calculate Heat Lost by Enzyme

The heat lost by the enzyme can be calculated using the formula: \[ Q = m imes c imes \Delta T \]where \( m = 0.016 \) kg (mass of the enzyme), \( c = 2250 \) J/kg·K (specific heat of the enzyme), and \( \Delta T = 19.5 \) K (change in temperature from 19.5°C to 0°C). Substituting the values, we get: \[ Q_{enzyme} = 0.016 \times 2250 \times 19.5 = 702 \, \text{J} \]
02

Calculate Heat Lost by Glass Vial

Similarly, calculate the heat lost by the glass vial:\[ Q = m \times c \times \Delta T \]where \( m = 0.006 \) kg (mass of the glass), \( c = 2800 \) J/kg·K (specific heat of the glass), and \( \Delta T = 19.5 \) K. Substituting the values, we get: \[ Q_{glass} = 0.006 \times 2800 \times 19.5 = 327.6 \, \text{J} \]
03

Total Heat Lost by System

The total heat lost by the system (enzyme and glass vial) is the sum of the heat lost by each component:\[ Q_{total} = Q_{enzyme} + Q_{glass} = 702 + 327.6 = 1029.6 \, \text{J} \]
04

Calculate Mass of Ice Melted

The heat required to melt ice is given by the latent heat equation:\[ Q = m \times L_f \]where \( m \) is the mass of the ice melted and \( L_f = 334,000 \) J/kg is the latent heat of fusion for ice. Setting the total heat lost equal to the heat required to melt the ice, we get:\[ 1029.6 = m \times 334,000 \]Solve for \( m \) to find the mass of ice melted:\[ m = \frac{1029.6}{334,000} \approx 0.00308 \text{ kg} \] or 3.08 g.

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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 an essential principle in thermodynamics. It's a measure of the amount of heat energy required to change the temperature of a substance by one degree.
In simple terms, it tells us how easily a substance can change its temperature. This quality depends on the substance's composition and its current state.
If a substance has a high specific heat capacity, it can absorb more heat before its temperature rises. This makes it ideal for regulating temperatures in various applications.To calculate it, we use the formula:
  • \( Q = m \times c \times \Delta T \)
  • \( Q \) is the heat absorbed or released,
  • \( m \) is the mass of the sample,
  • \( c \) is specific heat capacity, and
  • \( \Delta T \) is the change in temperature.
This calculation helps us understand energy transfers and find how much heat is needed for specific temperature changes. Applying this to different substances, like the enzyme and glass in our exercise, allows us to see how each material contributes to the total heat changes in a system.
Latent Heat of Fusion
The latent heat of fusion is another core concept in thermodynamics. When a substance changes its state, such as from solid to liquid, it requires a specified amount of energy.
This energy change happens at a constant temperature and without changing the temperature of the substance.
The energy required for such a phase change is referred to as latent heat.The latent heat of fusion specifically refers to the energy needed to convert a solid into a liquid. For ice, this conversion requires 334,000 J/kg. Using the formula:
  • \( Q = m \times L_f \)
  • \( Q \) is the heat energy,
  • \( m \) is the mass of the solid or liquid changing state,
  • \( L_f \) is the latent heat of fusion.
In our exercise, once we calculated the total heat lost by the enzyme and glass vial, we needed this amount to find how much ice could melt. This principle is crucial in understanding real-world phenomena, such as why ice melts in a drink only slowly controls how fast the drink gets cold.
Heat Transfer
Heat transfer is a fundamental concept in thermodynamics, detailing the movement of heat energy between different substances.
Heat energy naturally flows from a hotter object to a colder one until thermal equilibrium is reached.
There are three major methods of heat transfer:
  • Conduction: Direct transfer of heat through contact between substances.
  • Convection: Transfer of heat through fluid movements, such as in liquids or gases.
  • Radiation: Transfer of heat through electromagnetic waves, without needing a medium.
In our specific exercise, heat is primarily being transferred through conduction. The enzyme and glass vial release their heat to the ice bath until reaching equilibrium. This lost heat is absorbed by the ice, causing it to melt, demonstrating the exchange of energy between the components of the system. Each step in understanding how heat moves helps us predict the behavior of substances in practical scenarios, from cooking to engineering applications.

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Most popular questions from this chapter

A carpenter builds an exterior house wall with a layer of wood 3.0 \(\mathrm{cm}\) thick on the outside and a layer of Styrofoam insulation 2.2 \(\mathrm{cm}\) thick on the inside wall surface. The wood has \(k=0.080 \mathrm{W} / \mathrm{m} \cdot \mathrm{K},\) and the Styrofoam has \(k=0.010 \mathrm{W} / \mathrm{m} \cdot \mathrm{K}\). The interior surface temperature is \(19.0^{\circ} \mathrm{C},\) and the exterior surface temperature is \(-10.0^{\circ} \mathrm{C}\) (a) What is the temperature at the plane where the wood meets the Styrofoam? (b) What is the rate of heat flow per square meter through this wall?

The rate at which radiant energy from the sun reaches the earth's upper atmosphere is about 1.50 \(\mathrm{kW} / \mathrm{m}^{2} .\) The distance from the earth to the sun is \(1.50 \times 10^{11} \mathrm{m},\) and the radius of the sun is \(6.96 \times 10^{8} \mathrm{m}\) . (a) What is the rate of radiation of energy per unit area from the sun's surface? (b) If the sun radiates as an ideal blackbody, what is the temperature of its surface?

An asteroid with a diameter of 10 \(\mathrm{km}\) and a mass of \(2.60 \times 10^{15} \mathrm{kg}\) impacts the earth at a speed of \(32.0 \mathrm{km} / \mathrm{s},\) landing in the Pacific Ocean. If 1.00\(\%\) of the asteroid's kinetic energy goes to boiling the ocean water (assume an initial water temperature of \(10.0^{\circ} \mathrm{C}\) ), what mass of water will be boiled away by the collision? (For comparison, the mass of water contained in Lake Superior is about \(2 \times 10^{15} \mathrm{kg} .\) )

(a) Calculate the one temperature at which Fahrenheit and Celsius thermometers agree with each other. (b) Calculate the one temperature at which Fahrenheit and Kelvin thermometers agree with each other.

A technician measures the specific heat of an unidentified liquid by immersing an electrical resistor in it. Electrical energy is converted to heat transferred to the liquid for 120 s at a constant rate of 65.0 W. The mass of the liquid is \(0.780 \mathrm{kg},\) and its temperature increases from \(18.55^{\circ} \mathrm{C}\) to \(22.54^{\circ} \mathrm{C}\) . (a) Find the average specific heat of the liquid in this temperature range. Assume that negligible heat is transferred to the container that holds the liquid and that no heat is lost to the surroundings. (b) Suppose that in this experiment heat transfer from the liquid to the container or surroundings cannot be ignored. Is the result calculated in part (a) an overestimate or an underestimate of the average specific heat? Explain.

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