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Assuming no heat losses, the hoat released by the condensation of \(x \mathrm{~g}\) of steam at \(100^{\circ} \mathrm{C}\) can be used to convert \(y \mathrm{~g}\) of ico at \(0^{\circ} \mathrm{C}\) into water at \(100^{\circ} \mathrm{C}\), the ratio \(x: \mathrm{y}\) is (a) \(1: 1\) (b) \(1: 2\) (c) \(1: 3\) (d) \(3: 1\)

Short Answer

Expert verified
The ratio is 1:3, so option (c) is correct.

Step by step solution

01

Understand the Problem

We have steam at 100°C condensing to water at 100°C, releasing heat, which will be absorbed by ice at 0°C to convert it to water at 100°C. We need to find the mass ratio of steam (\(x\)) to ice (\(y\)) for this process with no heat loss.
02

Heat Released by Condensing Steam

The heat released when 1 gram of steam condenses to water at 100°C is given by the latent heat of vaporization of water, which is 540 cal/g. Therefore, for \(x\) grams of steam, the heat released is \(540x\) cal.
03

Heat Absorbed by Ice to Become Water at 100°C

For ice, we first melt it using the latent heat of fusion, which is 80 cal/g. Then, we raise the temperature of the resulting water from 0°C to 100°C using the specific heat capacity of water, which is 1 cal/g°C. Thus, the total heat needed is \(80y + 100y = 180y\) cal for \(y\) grams of ice.
04

Set Up the Equation

Since the heat lost by the steam equals the heat gained by the ice: \(540x = 180y\). Simplify this equation to find \(x:y\).
05

Solve for the Ratio

Divide both sides by 180 to solve for the ratio: \(\frac{540x}{180} = \frac{180y}{180}\). This gives \(3x = y\) or \(x : y = 1 : 3\).
06

Conclusion

The ratio of the mass of steam to ice is \(1:3\). This confirms option (c) is correct.

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Key Concepts

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

Latent Heat
Latent heat is a fascinating concept in the world of heat transfer. It refers to the amount of heat energy required to change the state of a substance without changing its temperature. This is unique because even though the substance absorbs or releases heat, its temperature remains constant during this phase transition.

In scientific terms, latent heat is associated with processes such as melting, freezing, boiling, and condensing. When a substance changes its state, like from solid to liquid or liquid to gas, it either absorbs or releases energy. This energy is termed as latent heat.
  • Latent Heat of Fusion: This is the heat energy needed to convert a solid into a liquid without a temperature change.
  • Latent Heat of Vaporization: This is the heat required for a liquid to become gas at constant temperature.
Each substance has its distinct values for latent heat, making these transitions a central aspect of thermodynamics and heat transfer.
Condensation
Condensation is a critical process in heat transfer, involving the change of a substance from a gas to a liquid state. It is the reverse process of vaporization. During condensation, heat is released to the surrounding environment, making it an exothermic process.

This concept is essential in many natural and technological processes. For instance, when steam condenses on a cold surface, it releases latent heat. This released heat can then be used for other processes, such as heating another substance. This is precisely the scenario in our original exercise where steam at 100°C condenses into water, releasing heat that transforms ice into water. This demonstrates how effective condensation is in transferring energy across states of matter.
Latent Heat of Vaporization
The latent heat of vaporization plays a crucial role in phase changes involving liquids and gases. For water, this is the energy required to change water from a liquid to a steam at its boiling point without a change in temperature. Specifically, for one gram of water, it requires 540 calories of energy to transition into steam.

In the context of our exercise, when steam condenses back into water at the same temperature, it releases this energy. Therefore, when you have steam at 100°C condensing, it liberates a substantial amount of heat energy. This energy can then be used for various applications, such as heating other substances or systems, effectively making it a crucial component in heat exchange systems.
Latent Heat of Fusion
The latent heat of fusion is the energy needed to change a solid into a liquid at constant temperature, specifically at the melting point. For ice, which is solid water, the latent heat of fusion is 80 calories per gram.

In our exercise, the latent heat of fusion is integral in melting ice at 0°C into liquid water. This transformation is the first step in the process of turning the ice into water at 100°C. After melting, additional heat is needed to increase the water temperature to 100°C, using the specific heat capacity of water. Together, these processes underline the importance of latent heat of fusion in the thermal dynamics of phase changes, allowing us to calculate and predict how much energy is required or released during these changes.

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Most popular questions from this chapter

The quantity of heat which crosses unit area of metal plate during conduction depends upon (a) Density of metal (b) Temperature gradient (c) Temperature to which metal is heated (d) Area of metal plate

A closely thermally insulated vessel contains \(100 \mathrm{~g}\) of water at \(0^{\circ} \mathrm{C} .\) If the air from this vessel is rapidly pumped out, intensive evaporation will produce cooling and as a result of this water frece . How much ice will be formed by this method? I" latent heat of fusion is \(80 \mathrm{cal} / \mathrm{g}\) and of cvaporation \(560 \mathrm{cal} / \mathrm{g}\).

In an experiment on the specilic heat of a metal a \(0.20 \mathrm{~kg}\) block of the metal at \(150^{\circ} \mathrm{C}\) is dropped in a copper calorimeter (of water equivalent \(0.025 \mathrm{~kg}\) ) containing \(150 \mathrm{~cm}^{3}\) of water at \(27^{\circ} \mathrm{C}\). The final temperature is \(40^{\circ} \mathrm{C}\). Calculalc the spocific heat ol the metal.

Statement-1 : Steam burns are more scvere those caused by boiling water while temperature of both steam and water is same i.c. \(100^{\circ} \mathrm{C}\). Statement-2: The internal cnergy of steam al \(100^{\circ} \mathrm{C}\). is much higher than internal energy of water al \(100^{\circ} \mathrm{C}\).

A body cools from 80 "C to 70 " 10 minutes. Find the time required further for it to cool from \(70^{\circ} \mathrm{C}\) to \(60^{\circ} \mathrm{C} .\) Assume the temperature of the surrounding to be \(30^{\circ} \mathrm{C}\).

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