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Chapter 15: Problem 55

In an energy recycling process, \(X g\) of steam at \(100^{\circ} \mathrm{C}\) becomes water at \(100^{\circ} \mathrm{C}\) which converts \(Y \mathrm{~g}\) of ice at \(0^{\circ} \mathrm{C}\) into water at \(100^{\circ} \mathrm{C}\). The ratio of \(X / Y\) will be : (a) \(\frac{1}{3}\) (b) \(\frac{2}{3}\) (c) 3 (d) 2

Short Answer

Expert verified
The ratio \(X/Y\) is 3, so the answer is (c).

Step by step solution

01

Understanding the Energy Exchange

In this process, we're dealing with two energy exchanges: the condensation of steam into water and the melting and subsequent heating of ice. These exchanges are governed by specific heat capacities and latent heats.
02

Energy Released by Steam

When steam at 100掳C condenses to water at 100掳C, it releases latent heat. The amount of energy (heat) released by the steam can be calculated as:\[ Q_{ ext{steam}} = X imes L_v \]Where \(L_v = 540 \text{ cal/g}\) is the latent heat of vaporization for water.
03

Energy Absorbed by Ice

The ice absorbs heat first to melt into water (latent heat of fusion) and then to heat the water from 0掳C to 100掳C. The energy absorbed is:\[ Q_{ ext{ice}} = Y imes L_f + Y imes C imes 100 \]Where \(L_f = 80 \text{ cal/g}\) is the latent heat of fusion for ice, and \(C = 1 \text{ cal/g掳C}\) is the specific heat capacity of water.
04

Setting the Energy Equation

Since all the energy released by the steam is absorbed by the ice, we can set up the equation:\[ X imes 540 = Y imes 80 + Y imes 1 imes 100 \]
05

Simplifying the Equation

Simplify and solve the energy equation:\[ 540 X = 80 Y + 100 Y \]\[ 540 X = 180 Y \]
06

Calculating the Ratio X/Y

To find the ratio \(X/Y\), divide both sides of the simplified equation by 180:\[ \frac{540}{180} = \frac{Y}{Y} \]\[ X/Y = 3 \]
07

Concluding the Solution

Thus, the ratio of \(X/Y\) is 3, which corresponds to option (c).

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Key Concepts

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

Latent Heat of Vaporization
The latent heat of vaporization refers to the amount of heat required to convert a unit mass of a liquid into vapor without a change in temperature. This energy facilitates the phase change from liquid to gas. For water, this constant amount of energy is known to be 540 cal/g.
When steam cools and condenses into water at the same temperature (100掳C), it releases a substantial amount of energy in the form of heat because of this latent heat of vaporization.
In practical scenarios, understanding the latent heat of vaporization helps us calculate the energy transformations during the condensation and boiling processes. This knowledge is vital for various applications, like designing efficient heating systems and understanding natural weather patterns.
Latent Heat of Fusion
The latent heat of fusion is the energy required to change a unit mass of a solid into liquid at its melting point. For ice, the latent heat of fusion is approximately 80 cal/g.
This property is crucial when ice melts into water without changing temperature. All the heat absorbed by the ice focuses on breaking the bonds holding the molecules in a solid structure.
During exercises like the one solved, knowing the latent heat of fusion lets us calculate how much energy is needed to convert ice at 0掳C to water at 0掳C. It sets the foundation for further heating processes and is essential in contexts such as climate studies and refrigeration.
Specific Heat Capacity
Specific heat capacity represents the amount of heat needed to raise the temperature of one gram of a substance by one degree Celsius. For water, this is 1 cal/g掳C. This property explains how substances absorb heat when they warm up.
In our problem, after the ice becomes water through fusion, it then absorbs more heat energy to rise from 0掳C to 100掳C. The water's specific heat capacity allows us to compute the energy required for this temperature change.
Understanding the specific heat capacity is vital because it determines how substances respond to heat and temperature changes. This concept applies universally from chemistry lab experiments to broader engineering and environmental calculations.

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Most popular questions from this chapter

At a temperature \(t^{\circ} \mathrm{C}\), a liquid is completely filled in a spherical shell of copper. If \(\Delta T\) increases temperature of the liquid and the shell, then the outward pressure \(d P\) on the shell resulted from increase in temperature is given by: (Given, \(K=\) Bulk modulus of the liquids, \(\gamma=\) coefficient of volume expansion, \(\alpha=\) coefficient of linear expansion of the material of the shell) (a) \(\frac{K}{2}(\gamma-3 \alpha) \Delta T\) (b) \(K(3 \alpha-\gamma) \Delta T\) (c) \(3 \alpha(K-\gamma) \Delta T\) (d) \(\gamma(3 \alpha-K) \Delta T\)

If in \(1.1 \mathrm{~kg}\) of water which is contained in a calorimeter of water equivalent \(0.02 \mathrm{~kg}\) at \(15^{\circ} \mathrm{C}\), steam at \(100^{\circ} \mathrm{C}\) is passed, till the temperature of calorimeter and its contents rises to \(80^{\circ} \mathrm{C}\). The mass of steam condensed in kilogram is: (a) \(0.131\) (b) \(0.065\) (c) \(0.260\) (d) \(0.135\)

\(5 \mathrm{~g}\) of water at \(30^{\circ} \mathrm{C}\) and \(5 \mathrm{~g}\) of ice at \(-20^{\circ} \mathrm{C}\) are mixed together in a calorimeter. The water equivalent of calorimeter is negligible and specific heat and latent heat of ice are \(0.5 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}\) and \(80 \mathrm{cal} / \mathrm{g}\) respectively. The final temperature of the mixure is: (a) \(0^{\circ} \mathrm{C}\) (b) \(-8^{\circ} \mathrm{C}\) (c) \(-4^{\circ} \mathrm{C}\) (d) \(2^{\circ} \mathrm{C}\)

Which of the following carries anomalous expansion ? (a) Mercury (b) Water (c) Copper (d) Sodium

It takes 20 minutes to melt \(10 \mathrm{~g}\) of ice, when rays from the 'sun are focussed by a lens of diameter \(5 \mathrm{~cm}\) on to a block of ice. The heat received from the sun on \(1 \mathrm{~cm}^{2}\) per minute is : (Given: \(L=80 \mathrm{k} \mathrm{cal} / \mathrm{kg}\) ) (a) \(R=2.04 \mathrm{cal} / \mathrm{cm}^{2}-\mathrm{min}\) (b) \(R=3.04 \mathrm{cal} / \mathrm{cm}^{2}-\mathrm{min}\) (c) \(R=0.204 \mathrm{cal} / \mathrm{cm}^{2}-\min\) (d) \(R=204 \mathrm{cal} / \mathrm{cm}^{2}-\mathrm{min}\)
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