/*! 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 52 Liquid nitrogen with a mass of \... [FREE SOLUTION] | 91Ó°ÊÓ

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Liquid nitrogen with a mass of \(100 \mathrm{g}\) at \(77.3 \mathrm{K}\) is stirred into a beaker containing \(200 \mathrm{g}\) of \(5.00^{\circ} \mathrm{C}\) water. If the nitrogen leaves the solution as soon as it turns to gas, how much water freezes? (The latent heat of vaporization of nitrogen is \(48.0 \mathrm{cal} / \mathrm{g},\) and the latent heat of fusion of water is \(79.6 \mathrm{cal} / \mathrm{g} .)\)

Short Answer

Expert verified
The amount of water that freezes is \(47.7 g\).

Step by step solution

01

Calculate Heat Absorbed by Nitrogen

First calculate the heat absorbed by the nitrogen when it turns into a gas. This is done by multiplying the mass of the nitrogen (\(100 \mathrm{g}\)) by the latent heat of vaporization of nitrogen (\(48.0 \mathrm{cal/g}\)):\n\n\(\Delta Q_1 = m_1 \cdot L_v = 100 \mathrm{g} \cdot 48.0 \mathrm{cal/g} = 4800 \mathrm{cal}\)
02

Calculate Heat Lost by Water

Secondly, calculate the heat lost by the water when it cools down from \(5.0^{\circ}\) to \(0.0^{\circ}\). The heat lost can be calculated using the formula \(\Delta Q = MC\Delta T\), where M is the mass of the water (\(200 \mathrm{g}\)), C is the specific heat capacity of water (\(1 \mathrm{cal/g^{\circ}C}\)), and \(\Delta T\) is the change in temperature (\(5.0^{\circ}C - 0.0^{\circ}C = 5.0^{\circ}C\)).\n\nSo, \(\Delta Q_2 = MC\Delta T = 200 \mathrm{g} \cdot 1 \mathrm{cal/g^{\circ}C} \cdot 5.0^{\circ}C = 1000 \mathrm{cal}\)
03

Find the Heat Used to Melt the Water

Now you know that the nitrogen absorbed \(4800 \mathrm{cal}\) and the water lost \(1000 \mathrm{cal}\), there is still \(4800 \mathrm{cal} - 1000 \mathrm{cal} = 3800 \mathrm{cal}\) that can be used to melt some of the water. The mass of the water that freezes can be found by using the formula \(\Delta Q = m \cdot L_f\), where \(\Delta Q\) is the heat used to melt the water, m is the mass, and \(L_f\) is the latent heat of fusion of water (\(79.6 \mathrm{cal/g}\)). So: \n\n\(3800 \mathrm{cal} = m \cdot 79.6 \mathrm{cal/g}\)\n\nSolving for m gives: \n\n\(m = \frac{3800 \mathrm{cal}}{79.6 \mathrm{cal/g}} = 47.7 \mathrm{g}\)
04

Conclusion

This is the amount of water that freezes. So, when \(100 g\) of nitrogen gas at \(77.3 K\) is stirred into \(200 g\) of water at \(5.0^{\circ}C\), \(47.7 g\) of the water freezes.

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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 one substance to another due to a temperature difference. It occurs naturally from a warmer object to a cooler one, until thermal equilibrium is reached. There are different modes of heat transfer including conduction, convection, and radiation. When dealing with phase changes and specific heat, conduction is often the focus
particularly in liquids and solids.
Understanding heat transfer is essential when analyzing the interaction between the liquid nitrogen and water in the exercise. The nitrogen absorbs heat from the warmer water, causing it to evaporate and the water to cool. This heat exchange process continues until the desired phase change occurs. This concept underlines all of the calculations and explanations in the provided solution.
Latent Heat
Latent heat is the heat absorbed or released during a phase change, without changing the substance's temperature. There are different kinds of latent heat, such as latent heat of fusion and latent heat of vaporization. In the given exercise:
  • **Latent Heat of Vaporization:** For nitrogen, this is 48.0 cal/g. This value tells us the quantity of heat needed to convert 1 gram of liquid nitrogen into gas without a temperature change.
  • **Latent Heat of Fusion:** For water, this is 79.6 cal/g. This indicates the heat needed to turn 1 gram of ice into water, or vice versa, without changing temperature.
Latent heat calculations help determine how much of a substance will freeze or melt, without requiring a change in temperature.
Phase Change
A phase change involves a transition of a substance from one state of matter to another, such as
from liquid to gas, or from water to ice. In the context of the exercise, nitrogen undergoes a phase change from liquid to gas, while for water, the change is from liquid to solid.
Here’s how these phase changes occur:
  • The liquid nitrogen absorbs sufficient heat to transition into a gaseous state, contributing to the cooling of the water.
  • As heat is removed from the liquid water, it freezes, indicating the change to a solid phase in the presence of diminishing heat energies such as the latent heat of fusion.
Understanding phase changes is crucial because they are intertwined with the energy changes essential to heat transfer calculations.
Specific Heat Capacity
Specific heat capacity refers to the amount of heat required to change a substance's temperature by one degree Celsius per unit mass. This property varies across different substances.
In the exercise:
  • For water, a familiar substance, the specific heat capacity is 1 cal/g°C. This means heating or cooling 1 gram of water by 1°C requires 1 calorie of energy.
The heat calculations involved in cooling water to its freezing point before a phase change, rely heavily on this concept. It demonstrates how much thermal energy per gram must be lost by the water to facilitate this cooling. By understanding concepts like specific heat capacity, we can better predict temperature changes resulting from energy exchanges during interactions between substances.

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