/*! 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 72 How much heat transfer (in kiloc... [FREE SOLUTION] | 91Ó°ÊÓ

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How much heat transfer (in kilocalories) is required to thaw a 0.450 -kg package of frozen vegetables originally at \(0^{\circ} \mathrm{C}\) if their heat of fusion is the same as that of water?

Short Answer

Expert verified
The heat transfer required to thaw a 0.450 kg package of frozen vegetables originally at 0°C is 36 kilocalories.

Step by step solution

01

Convert mass of vegetables to grams

To calculate the heat transfer, we need the mass of the vegetables in grams. Given the mass is 0.450 kg: 1 kg = 1000 g 0.450 kg = 0.450 * 1000 g = 450 g The mass of the package of frozen vegetables is 450 g.
02

Calculate the heat transfer using heat of fusion formula

Now that we have the mass in grams, we can use the heat of fusion formula: Heat transfer (Q) = Mass (m) × Heat of fusion (Lf) Heat of fusion of water (Lf) = 80 cal/g Putting the values in the formula, we get: Q = m × Lf Q = 450 g × 80 cal/g
03

Calculate the heat transfer

Multiply the mass with the heat of fusion: Q = 450 g × 80 cal/g = 36000 cal So, the heat transfer required is 36000 calories.
04

Convert calories to kilocalories

Lastly, let's convert the heat transfer from calories to kilocalories. To do that: 1 kcal = 1000 cal 36000 cal = 36000/1000 kcal = 36 kcal The heat transfer required to thaw a 0.450 kg package of frozen vegetables originally at 0°C is 36 kilocalories.

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Key Concepts

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

Heat of Fusion
The heat of fusion is the amount of energy needed to change a substance from the solid phase to the liquid phase at its melting point without changing its temperature. This property is crucial when calculating the energy required to melt a frozen object, such as a package of vegetables, at its melting point, which for water and water-rich substances like vegetables is typically 0°C.

For water, the heat of fusion is approximately 80 calories per gram. This means that to melt 1 gram of ice (or in our context, frozen vegetables) at 0°C, we need to supply 80 calories of energy. When solving the exercise, we applied this concept by multiplying the mass of the vegetables by the heat of fusion to find the total heat transfer needed.
Thermodynamics

The field of thermodynamics deals with the principles of heat and temperature and their relation to energy and work. Fundamental to thermodynamics is the concept of energy conservation, often referred to as the First Law of Thermodynamics. It states that the total energy of an isolated system is constant; energy can neither be created nor destroyed, only converted from one form to another.


When we discuss heat transfer, such as the process of thawing frozen vegetables, we're examining a practical application of thermodynamic principles. The exercise illustrates how thermal energy is transferred to the frozen vegetables to trigger a phase change, fully aligning with the thermodynamic concept of energy conversion.

Calorimetry
Calorimetry is the science of measuring the amount of heat. All calorimetric techniques are therefore based on the measurement of heat that can be detected during a physical or chemical process. In our exercise, when calculating the amount of heat needed to thaw the vegetables, we used a simple calorimetric concept, assuming the heat of fusion provided was an empirical value obtained from calorimetric measurements.

During such measurements, any heat exchanged with the surroundings is carefully accounted for, which in laboratory settings is done using an instrument called a calorimeter. In more complex exercises, you could be asked to account for specific heat capacities, heat losses to the environment, or the heat required to raise the temperature of a substance, all of which fall under the domain of calorimetry.
Energy Conversion
The concept of energy conversion highlights the transformation of energy from one form into another. In thermodynamics, this is a fundamental process that occurs naturally in various systems. In the case of melting ice or thawing frozen vegetables, the energy needed for the heat of fusion comes from the surroundings, often in the form of thermal energy, and is converted into the latent heat absorbed during the melting process.

This energy does not raise the temperature of the vegetables but rather breaks the molecular bonds within the solid, allowing it to change state. The term 'latent' refers to the fact that this heat is 'hidden', as it does not result in a temperature change that can be felt or measured by a conventional thermometer.

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Most popular questions from this chapter

(a) Calculate the rate of heat conduction through house walls that are \(13.0 \mathrm{cm}\) thick and have an average themal conductivity twice that of glass wool. Assume there are no windows or doors. The walls' surface area is \(120 \mathrm{m}^{2}\) and their inside surface is at \(18.0^{\circ} \mathrm{C}\), while their outside surface is at \(5.00^{\circ} \mathrm{C} .\) (b) How many \(1-\mathrm{k} \mathrm{W}\) room heaters would be needed to balance the heat transfer due to conduction?

Describe a situation in which heat transfer occurs.

On a trip, you notice that a 3.50 -kg bag of ice lasts an average of one day in your cooler. What is the average power in watts entering the ice if it starts at \(0^{\circ} \mathrm{C}\) and completely melts to \(0^{\circ} \mathrm{C}\) water in exactly one day?

Most cars have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made of copper and is filled to its 16.0 -L capacity when at \(10.0^{\circ} \mathrm{C}\). What volume of radiator fluid will overflow when the radiator and fluid reach a temperature of \(95.0^{\circ} \mathrm{C},\) given that the fluid's volume coefficient of expansion is \(\beta=400 \times 10^{-6} /^{\circ} \mathrm{C} ?\) (Your answer will be a conservative estimate, as most car radiators have operating temperatures greater than \(95.0^{\circ} \mathrm{C}\) ).

(a) Determine the power of radiation from the Sun by noting that the intensity of the radiation at the distance of Earth is \(1370 \mathrm{W} / \mathrm{m}^{2}\). Hint: That intensity will be found everywhere on a spherical surface with radius equal to that of Earth's orbit. (b) Assuming that the Sun's temperature is 5780 K and that its emissivity is 1, find its radius.

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