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Chapter 18: Problem 15

\(1 \mathrm{~g}\) of ice at \(0^{\circ} \mathrm{C}\) is mixed with \(1 \mathrm{~g}\) of steam at \(100^{\circ} \mathrm{C}\). After thermal equilibrium is attained, the temperature of mixture will be : (a) \(1^{\circ} \mathrm{C}\) (b) \(50^{\circ} \mathrm{C}\) (c) \(81^{\circ} \mathrm{C}\) (d) \(100^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The final temperature is 100掳C.

Step by step solution

01

Understand the Problem

We need to find the final equilibrium temperature of a mixture resulting from 1 g of ice at 0掳C and 1 g of steam at 100掳C.
02

Calculate Energy to Melt Ice

The latent heat of fusion for ice is 334 J/g. To convert 1 g of ice at 0掳C to water at 0掳C, the energy required is calculated as follows:\[ Q_1 = m_i \times L_f = 1 \, \text{g} \times 334 \, \text{J/g} = 334 \, \text{J} \] where \(m_i\) is the mass of ice and \(L_f\) is the latent heat of fusion.
03

Calculate Energy to Condense Steam

The latent heat of vaporization for steam is 2260 J/g. To convert 1 g of steam at 100掳C to water at 100掳C, the energy released is:\[ Q_2 = m_s \times L_v = 1 \, \text{g} \times 2260 \, \text{J/g} = 2260 \, \text{J} \] where \(m_s\) is the mass of steam and \(L_v\) is the latent heat of vaporization.
04

Compare Energy for Phase Changes

The steam releases 2260 J as it condenses to water, whereas the ice requires only 334 J to melt into water. Since 2260 J is more than 334 J, all the ice will melt, and there will still be leftover energy.
05

Calculate Remaining Energy

Subtract the energy used to melt the ice from the energy released by the steam:\[ Q_{\text{left}} = Q_2 - Q_1 = 2260 \, \text{J} - 334 \, \text{J} = 1926 \, \text{J} \]This is the energy available to raise the temperature of 2 g of water (1 g from melted ice and 1 g from condensed steam).
06

Calculate Final Temperature of Water

Use the specific heat capacity of water (4.18 J/g掳C) to find the temperature rise:\[ \Delta T = \frac{Q_{\text{left}}}{m_t \times c} = \frac{1926 \, \text{J}}{2 \, \text{g} \times 4.18 \, \text{J/g掳C}} \approx 230.38 \, \text{掳C} \]Add this to the initial temperature of water at 0掳C (after ice melts) to get the final temperature, limited by the boiling point:\[ T_f = 100 \, \text{掳C} \] since water cannot exceed 100掳C at standard pressure.

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Key Concepts

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

Latent Heat of Fusion
When substances change from one phase to another, such as ice turning into liquid water, they require a certain amount of energy, even if there is no change in temperature. This energy is called the latent heat of fusion. It is the energy needed to change a substance from solid to liquid at its melting point. For ice, this is a significant amount, specifically 334 J/g. This means it takes 334 Joules to melt 1 gram of ice at 0掳C into water at 0掳C. The energy goes into breaking the bonds between the molecules in the solid state, transforming them into the more free-moving liquid state.

In the problem, the entire energy needed to turn the ice into water is provided by the steam condensing at the higher energy level. Therefore, the ice does not need any external temperature increase to achieve its phase change.
Latent Heat of Vaporization
The process of converting steam (gas) into liquid also involves a significant amount of energy, but this time the energy is released. This is known as the latent heat of vaporization. It is the energy released when one gram of a substance condenses from gas to liquid. For water, this value is quite high: 2260 J/g. This energy results from the vigorous vibrations and motions of water molecules in the gas phase being compressed into the less energetic liquid phase.

In this exercise, as 1 g of steam condenses, it releases 2260 Joules of energy which is much more than the 334 Joules required for the ice to melt. This excess energy is then used to increase the overall temperature of the resultant water mixture, after the phase changes have been completed.
Specific Heat Capacity
Specific heat capacity is a measure of how much energy is required to raise the temperature of one gram of a substance by 1掳C. For water, this value is 4.18 J/g掳C. This is relatively high, meaning water needs a lot of energy to change its temperature, reflecting its ability to store heat.

After all the phase changes in the problem occur, there are 2 grams of water present. The excess energy from the steam's condensation is then utilized to increase the temperature of this water mixture. By using the specific heat capacity, we can predict how much the temperature will rise based on the remaining energy after the phase changes are complete.
Phase Change Energy Calculation
The exercise combines all these concepts to calculate the final temperature after the phase changes. First, the energy required to melt the ice (using latent heat of fusion) and the energy released to condense the steam (using latent heat of vaporization) are computed. The difference in these energies gives you the excess energy, which plays a crucial role.

This excess energy is then used to heat the resultant water to its equilibrium temperature. The specific heat capacity is applied here to determine how much the temperature of the resultant water increases. Calculations like these are essential for understanding thermal dynamics, ensuring you know each phase's specific transactions of energy to determine the end result accurately.

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Most popular questions from this chapter

Read the following statements carefully (A) Black body radiation is white (B) Emissivity of a body is equal to its absorptive power Mark correct option: (a) Statement (A) is correct

In a closed room, heat transfer takes place by: (a) conduction (b) convection (c) radiation (d) all of these

In a composite rod, when two rods of different lengths and of the same area of cross-section, are joined end to end then if \(K\) is the effective coefficient of thermal conductivity \(\frac{l_{1}+l_{2}}{K}\) is equal to : (a) \(\frac{l_{1}}{K_{1}}-\frac{l_{2}}{K_{2}}\) (b) \(\frac{l_{1}}{K_{2}}=\frac{l_{2}}{K_{1}}\) (c) \(\frac{l_{1}}{K_{1}}+\frac{l_{2}}{K_{2}}\) (d) \(\frac{l_{1}}{K_{2}}+\frac{l_{2}}{K_{1}}\)

The thermal conductivity of a material of a cylinder of radius \(R\) is \(K_{1}\), which is surrounded by a cylindrical shell of inner radius \(R\) and outer radius \(2 R\) made of the material of thermal conductivity \(K_{2}\). The two ends of the combined system are maintained at two different temperatures. Assume, that there is no loss of heat across the cylindrical surface and the system is in steady state. The effective thermal conductivity of the system is: (a) \(K_{1}+K_{2}\) (b) \(\frac{K_{1} K_{2}}{\left(K_{1}+K_{2}\right)}\) (c) \(\frac{\left(K_{1}+3 K_{2}\right)}{4}\) (d) \(\frac{\left(3 K_{1}+K_{2}\right)}{4}\)

A metal rod of length \(L\) and cross-sectional area \(A\) converts a large tank of water at temperature \(\theta_{0}\) and a small vessel containing mass \(m\) of water at initial temperature of \(\theta_{1}\left(<\theta_{0}\right)\). If the thermal conductivity of rod is \(K\), then the time taken for the temperature of water in smaller vessel to become \(\theta_{2}\left(\theta_{1}<\theta_{2}<\theta_{0}\right)\) is : (Given: Specific heat capacity of water is \(s\) and all other heat capacities are neglected) (矛) \(\bar{i}=\frac{m s}{K A} \operatorname{iog} \frac{\left(\theta_{0}-\theta_{1}\right)}{\left(\theta_{0}-\theta_{2}\right)}\) (b) \(T=\frac{L m s}{K A} \log \left(\frac{\theta_{0}-\theta_{1}}{\theta_{0}-\theta_{2}}\right)\) (c) \(T=\frac{2 m L s}{K A}\left(\log \frac{\left(\theta_{0}-\theta_{1}\right)}{\theta_{0}-\theta_{2}}\right)\) (d) none of the above
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