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(II) A 28-g ice cube at its melting point is dropped into an insulated container of liquid nitrogen. How much nitrogen evaporates if it is at its boiling point of 77 K and has a latent heat of vaporization of 200 kJ/kg? Assume for simplicity that the specific heat of ice is a constant and is equal to its value near its melting point.

Short Answer

Expert verified
46.76 g of nitrogen evaporates.

Step by step solution

01

Understand the Problem

We need to find out how much nitrogen evaporates when a 28-g ice cube at its melting point is added to liquid nitrogen at its boiling point.
02

Identify Known Values

The mass of the ice cube, \( m_{\text{ice}} = 28 \text{ g} = 0.028 \text{ kg} \), the latent heat of fusion of ice, \( L_f = 334 \text{ kJ/kg} \), and the latent heat of vaporization of nitrogen, \( L_v = 200 \text{ kJ/kg} \).
03

Calculate Energy Needed to Melt the Ice

Use the formula for latent heat: \( Q = m \cdot L \) to calculate the energy needed to melt the ice. For the ice, \( Q_{\text{melt}} = 0.028 \text{ kg} \times 334 \text{ kJ/kg} = 9.352 \text{ kJ} \).
04

Calculate the Mass of Evaporated Nitrogen

The energy to melt the ice is provided by the evaporating nitrogen. Using the formula \( Q = m_{\text{nitrogen}} \cdot L_v \), solve for the mass of nitrogen, \( m_{\text{nitrogen}} = \frac{Q_{\text{melt}}}{L_v} = \frac{9.352 \text{ kJ}}{200 \text{ kJ/kg}} = 0.04676 \text{ kg} \).
05

Convert Mass of Evaporated Nitrogen to Grams

Convert the mass of evaporated nitrogen from kilograms to grams: \( 0.04676 \text{ kg} \times 1000 \text{ g/kg} = 46.76 \text{ g} \).

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Key Concepts

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

Ice Melting
When we talk about ice melting, we're referring to the process of a solid turning into a liquid. For ice, this means it transforms into water. Melting occurs when ice absorbs enough energy (usually in the form of heat) to break the bonds holding its molecules in a rigid structure. At this crucial point, known as the melting point, the temperature doesn't rise until the whole ice cube has melted.

To melt ice, we use a specific amount of energy called the latent heat of fusion. The latent heat of fusion for ice is 334 kJ/kg. This means for every kilogram of ice that needs to be melted, 334 kilojoules of energy is required. This energy transfer must overcome the molecular forces holding the ice in place.
  • Key Point: Melting of ice stays at constant temperature until complete.
  • Energy Needed: Determined by latent heat of fusion.
Latent Heat of Vaporization
Latent heat of vaporization refers to the energy required to change a substance from a liquid to a gas without changing its temperature. For liquid nitrogen, this energy amount is particularly important given its common use in laboratories and industrial applications.

Liquid nitrogen boils at an extremely low temperature of 77 K. To evaporate and become gaseous nitrogen, it must absorb energy equivalent to its latent heat of vaporization, which is 200 kJ/kg. This energy facilitates the breaking of intermolecular attractions in the liquid, allowing the molecules to spread out and form a gas.
  • Purpose: To provide energy sufficient for phase transition without temperature change.
  • Application: Used in understanding energy exchanges in phenomena like evaporation.
Thermal Physics
In thermal physics, we study the theory of heat transfer and the change of energy from one form to another. This is crucial when dealing with processes like melting and vaporization, which involve large energy changes at constant temperatures.

The balance of energy within a system is a key concept here. For instance, when we say the energy used to melt ice is exactly the energy taken from the evaporated liquid nitrogen, we're observing the law of energy conservation. No energy is lost within the system; it's simply transformed.

Aspects like specific heat, latent heat, temperature change and heat capacity are fundamental. Each plays a role in determining how different substances react to heat addition or loss.
  • Core Idea: Energy balance and transformation.
  • Focus: Specific heat, latent heat, phase transition.

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Most popular questions from this chapter

(II) A 31.5-g glass thermometer reads 23.6\(^\circ\)C before it is placed in 135 mL of water. When the water and thermometer come to equilibrium, the thermometer reads 41.8\(^\circ\)C. What was the original temperature of the water? Ignore the mass of fluid inside the glass thermometer.

To get an idea of how much thermal energy is contained in the world's oceans, estimate the heat liberated when a cube of ocean water, 1 km on each side, is cooled by 1K. (Approximate the ocean water as pure water for this estimate.)

(II) A 215-g sample of a substance is heated to 330\(^\circ\)C and then plunged into a 105-g aluminum calorimeter cup containing 185 g of water and a 17-g glass thermometer at 10.5\(^\circ\)C. The final temperature is 35.0\(^\circ\)C. What is the specific heat of the substance? (Assume no water boils away.)

(\(a\)) Estimate the total power radiated into space by the Sun, assuming it to be a perfect emitter at \(T =\) 5500 K. The Sun's radius is \(7.0 \times 10^8\) m. (\(b\)) From this, determine the power per unit area arriving at the Earth, \(1.5 \times 10^{11}\) m away (Fig. 14\(-\)20).

(II) The \(heat \, capacity\), \(C\), of an object is defined as the amount of heat needed to raise its temperature by 1 C\(^\circ\). Thus, to raise the temperature by requires heat \(Q\) given by \(\Delta T\) $$Q = C \, \Delta \, T$$ (\(a\)) Write the heat capacity C in terms of the specific heat, \(c\), of the material. (\(b\)) What is the heat capacity of 1.0 kg of water? (\(c\)) Of 45 kg of water?

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