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A. \(100-\mathrm{g}\) 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 Tf, which is calculated following the steps above. Substituting the known values and solving the equation will give the final answer.

Step by step solution

01

Calculation of heat required to melt the ice

Calculate the total heat required to melt the ice cube. The heat, Q, required to melt a mass m of a substance, given its heat of fusion, Lf, is given by the equation Q=m*Lf. For ice, the heat of fusion is \[334 kJ/kg\]. Hence, the heat required to melt the 100g (or 0.1kg) ice cube is Q = 0.1kg * 334 kJ/kg = 33.4 kJ.
02

Calculation of heat lost by the water

Calculate the heat lost by 1.0 kg of water initially at 80°C as it cools down to the final temperature Tf. The heat lost, Q, by a mass m of substance as it cools down a temperature change (Delta T), given its specific heat capacity, c, is given by the equation Q=mc*Delta T. The specific heat capacity of water c=4.18 kJ/kg.K. Hence, the heat lost by the water as it cools down to Tf is Q = 1.0 kg * 4.18kJ/kg.K * (80°C - Tf).
03

Setting up the heat balance equation

Because no other heat transfers are taking place and the system is isolated, the heat lost by the water must be equal to the heat gained by the ice. Therefore, the heat required to melt the ice (Step 1) plus the heat required to warm the melted ice (water came from ice) to the final temperature have to equal the heat lost by the water (Step 2). The equation {0.1kg * 4.18kJ/kg.K * (Tf - 0°C) + 33.4 kJ = 1.0 kg * 4.18kJ/kg.K * (80°C - Tf)} should therefore be solved for Tf.
04

Solve for the final temperature

Rearrange and simplify the equation from step 3 to solve for Tf. This gives: {0.418 kJ/K * Tf + 33.4 kJ = 4.18 kJ/K * 80°C - 4.18 kJ/K * Tf}. After combining like terms and rearranging, find the value for Tf.

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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 thermal energy moves from one object or substance to another. It occurs whenever there is a temperature difference and continues until the objects reach thermal equilibrium, meaning they are at the same temperature.

There are three main modes of heat transfer:
  • Conduction, which occurs in solids where heat is transferred through direct contact.
  • Convection, which occurs in fluids (liquids or gases) where heat is carried through the fluid's movement.
  • Radiation, where heat is transferred through electromagnetic waves without requiring a medium.
In the exercise, heat transfer occurs when the ice cube is placed in the hot water. The heat moves from the warmer water to the colder ice until they reach the same temperature. Calculations, such as the one in the solution, allow us to find the amount of heat transferred and the new equilibrium temperature.
Specific Heat Capacity
Specific heat capacity is a property of a material that indicates how much heat energy is required to raise the temperature of a unit mass of the substance by one degree. It is usually denoted by the symbol 'c' and is measured in units like joules per kilogram per kelvin (J/kg·K) or kilojoules per kilogram per Kelvin (kJ/kg·K).

In the context of the exercise, water has a specific heat capacity of \(4.18 kJ/kg.K\). This means that to increase the temperature of one kilogram of water by one degree Celsius, \(4.18 kJ\) of heat is required. The high specific heat capacity of water compared to most other substances explains why water is effective in moderating temperature changes and why it takes a significant amount of heat to change its temperature.
Heat of Fusion
The heat of fusion, often symbolized as \(L_f\), is the amount of heat energy required to change a unit mass of a substance from solid to liquid state without changing its temperature. For water, the heat of fusion is approximately \(334 kJ/kg\).

This is a crucial concept in understanding the melting of ice in our exercise. The 100-gram ice cube needs a certain amount of heat to transition from solid ice to liquid water at the same temperature (\(0^\circ C\)). This phase change requires a substantial amount of energy due to the breaking of hydrogen bonds within the ice structure. The provided solution shows how this heat of fusion is used to calculate the total heat required to melt the ice, and that is a fundamental step in determining the final equilibrium temperature of the combined water and melted ice system.

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Most popular questions from this chapter

18 Overall, \(80 \%\) of the energy used by the body must be eliminated as excess thermal energy and needs to be dissipated. The mechanisms of elimination are radiation. evaporation of sweat (2 \(430 \mathrm{~kg} / \mathrm{kg})\), evaporation from the hangs \((38 \mathrm{~kJ} / \mathrm{h})\), conduction, and convection. A person working out in a gym has a metabolic rate of \(2500 \mathrm{~kJ} / \mathrm{h} .\) His body temperature is \(37^{\circ} \mathrm{C}\), and the outside temperature \(24^{\circ} \mathrm{C}\). Assume the skin has an area of \(2.0 \mathrm{~m}^{2}\) and cmissivity of \(0.97\). (a) At what rate is his excess thermal energy dissipated by radiation? (b) If he eliminates \(0.40 \mathrm{~kg}\) of perspiration during that hour, at what rate is thermal energy dissipated by evaporation of sweat? (c) AL what rate is energy eliminated by evaporation from the lungs? (d) At what rate must the remaining excess energy be eliminated through conduction and convection?

Calculate the temperature at which a tungsten filament that has an emissivity of \(0.90\) and a surface area of \(2.5 \times 10^{-5} \mathrm{~m}^{2}\) will radiate energy at the rate of \(25 \mathrm{~W}\) in a room where the temperature is \(22^{\circ} \mathrm{C}\).

A \(50-g\) ice cube at \(0^{\circ} \mathrm{C}\) is heated until \(45 \mathrm{~g}\) has become water at \(100^{\circ} \mathrm{C}\) and \(5.0 \mathrm{~g}\) has become steam at \(100^{\circ} \mathrm{C}\). How much energy was added to accomplish the transformation?

In a showdown on the streets of Laredo, the good guy drops a \(5.0-g\) silver bullet at a temperature of \(20^{\circ} \mathrm{C}\) into a \(100-\mathrm{cm}^{3}\) cup of water at \(90^{\circ} \mathrm{C}\). Simultaneously, the bad guy drops a \(5.0-\mathrm{g}\) copper bullet at the same initial temperature into an identical cup of water. Which one ends the showdown with the coolest cup of water in the West? Neglect any energy transfer into or away from the container.

A "solar cooker" consists of a curved reflecting mirror that focuses sumlight onto the object to be heated (Fig. \(P 11.67\) ). The solar power per unit area reaching the Farth at the location of a \(0.50-\mathrm{m}-\) diameter solar cooker is \(600 \mathrm{~W} / \mathrm{m}^{2}\). Assuming \(50 \%\) of the incident energy is converted to thermal energy, how long would it take to boil away \(1.0 \mathrm{~L}\) of water initially at \(20^{\circ} \mathrm{C} ?\) (Neglect the specific heat of the container.)

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