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Chapter 11: Problem 27

A \(100-g\) cube of ice at \(0^{\circ} \mathrm{C}\) is dropped into \(1.0 \mathrm{~kg}\) of water that was originally at \(80^{\circ} \mathrm{C}\). What is the final temperature of the water after the ice has melted?

Short Answer

Expert verified
The final temperature of the water after the ice has melted is \(65.4 \, ^{\circ} \mathrm{C}\).

Step by step solution

01

Calculate the heat lost by the water

First, calculate the amount of heat (\(Q\)) that the water will lose as it cools down to \(0^{\circ} \mathrm{C}\). Use the formula \(Q = mc\Delta T\), where \(m = 1.0 \,kg\) is the mass of the water, \(\Delta T = 80 \, ^{\circ} \mathrm{C} - 0 \, ^{\circ} \mathrm{C} = 80 \, ^{\circ} \mathrm{C}\) is the change in temperature, and \(C = 4.18 \, kJ/kg \cdot ^{\circ} \mathrm{C}\) is the specific heat capacity of water. Thus, \(Q = 1.0 \, kg * 4.18 \, kJ/kg \cdot ^{\circ} \mathrm{C} * 80 \, ^{\circ} \mathrm{C} = 334.4 \, kJ\).
02

Calculate the heat absorbed by the ice

Next, calculate the amount of heat (Q) that the ice will absorb as it melts and warms up to \(0^{\circ} \mathrm{C}\). Use the formula \(Q = mL\), where \(m = 0.1 \, kg\) is the mass of the ice and \(L = 333.5 \, kJ/kg\) is the heat of fusion of ice. Thus, \(Q = 0.1 \, kg * 333.5 \, kJ/kg = 33.35 \, kJ\).
03

Find the final temperature of the water

Lastly, subtract the heat absorbed by the ice from the heat lost by the water (334.4 \, kJ - 33.35 \, kJ = 301.05 \, kJ) and divide by the total mass of the water and melted ice (1.0 kg + 0.1 kg) and the specific heat capacity of water. This will give the final temperature of the water in 掳C: \(T_{\mathrm{final}} = \frac{Q}{mc} = \frac{301.05 \, kJ}{1.1 \, kg * 4.18 \, kJ/kg\cdot ^{\circ} \mathrm{C}} = 65.4 \, ^{\circ} \mathrm{C}\). This means that the final temperature of the water after the ice has melted is \(65.4 \, ^{\circ} \mathrm{C}\).

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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 by which heat energy moves from a warmer object to a cooler one. It's an everyday phenomenon you've witnessed when you let a hot cup of coffee cool down or when ice melts in your drink. There are three main methods of heat transfer: conduction, convection, and radiation. In our exercise, conduction plays a role as the ice absorbs heat from the surrounding water.

During the heat transfer process, no heat is lost to the environment; it simply moves from the hot water to the colder ice. The objective in thermodynamics exercises is often to find out how this transfer changes both substances. The exercise provided shows a practical application of this concept, where the ice cube reaches thermal equilibrium with the warm water, resulting in a temperature change for both.
Specific Heat Capacity
The specific heat capacity is a property that indicates how much heat energy is required to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). Different substances have different capacities to store heat, which is why they heat up or cool down at varying rates.

In our exercise, water has a specific heat capacity of \( 4.18 \, kJ/kg \cdot ^{\circ} \mathrm{C} \). This means for every kilogram of water to increase temperature by \(1^{\circ} \mathrm{C}\), it requires 4.18 kJ of heat energy. This concept is crucial to calculating the energy involved in heating or cooling a substance within thermodynamic systems.
Heat of Fusion
The amount of heat required to convert a solid at its melting point to a liquid without an increase in temperature is known as the heat of fusion. This is a unique value for every substance, reflecting the amount of energy needed to overcome the forces keeping its particles in a rigid structure.

In the exercise solution, the heat of fusion for ice is given as \(333.5 \, kJ/kg\), signifying the energy needed to melt each kilogram of ice at \(0^{\circ} \mathrm{C}\) without changing its temperature. Knowledge of the heat of fusion is vital when calculating the energy exchange involved in phase changes, such as turning ice into water.

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Most popular questions from this chapter

The highest recorded waterfall in the world is found at Angel Falls in Venezuela. Its longest single waterfall has a height of \(807 \mathrm{~m}\). If water at the top of the falls is at \(15.0^{\circ} \mathrm{C}\), what is the maximum temperature of the water at the bottom of the falls? Assume all the kinetic energy of the water as it reaches the bottom goes into raising the water's temperature.

A steam pipe is covered with \(1.50-\mathrm{cm}\)-thick insulating material of thermal conductivity \(0.200 \mathrm{cal} / \mathrm{cm} \cdot{ }^{\circ} \mathrm{C} \cdot \mathrm{s}\). How much energy is lost every second when the steam is at \(200^{\circ} \mathrm{C}\) and the surrounding air is at \(20.0^{\circ} \mathrm{C}\) ? The pipe has a circumference of \(800 \mathrm{~cm}\) and a length of \(50.0 \mathrm{~m}\). Neglect losses through the ends of the pipe.

A class of 10 students taking an exam has a power output per student of about \(200 \mathrm{~W}\). Assume the initial temperature of the room is \(20^{\circ} \mathrm{C}\) and that its dimensions are \(6.0 \mathrm{~m}\) by \(15.0 \mathrm{~m}\) by \(3.0 \mathrm{~m}\). What is the temperature of the room at the end of \(1.0 \mathrm{~h}\) if all the energy remains in the air in the room and none is added by an outside source? The specific heat of air is \(837 \mathrm{~J} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), and its density is about \(1.3 \times 10^{-3} \mathrm{~g} / \mathrm{cm}^{3}\).

A Styrofoam box has a surface area of \(0.80 \mathrm{~m}^{2}\) and a wall thickness of \(2.0 \mathrm{~cm}\). The temperature of the inner surface is \(5.0^{\circ} \mathrm{C}\), and the outside temperature is \(25^{\circ} \mathrm{C}\). If it takes \(8.0 \mathrm{~h}\) for \(5.0 \mathrm{~kg}\) of ice to melt in the container, determine the thermal conductivity of the Styrofoam.

Three liquids are at temperatures of \(10^{\circ} \mathrm{C}, 20^{\circ} \mathrm{C}\), and \(30^{\circ} \mathrm{C}\), respectively. Equal masses of the first two liquids are mixed, and the equilibrium temperature is \(17^{\circ} \mathrm{C}\). Equal masses of the second and third are then mixed, and the equilibrium temperature is \(28^{\circ} \mathrm{C}\). Find the equilibrium temperature when equal masses of the first and third are mixed.
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