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A 4.00-kg silver ingot is taken from a furnace, where its temperature is 750.0\(^\circ\)C, and placed on a large block of ice at 0.0\(^\circ\)C. Assuming that all the heat given up by the silver is used to melt the ice, how much ice is melted?

Short Answer

Expert verified
Approximately 2111 grams of ice is melted.

Step by step solution

01

Understand the Energy Transfer

The heat lost by the silver ingot will be absorbed by the ice, causing the ice to melt. We use the formula for heat transfer: \[ q = m \cdot c \cdot \Delta T \] where \( q \) is the heat energy, \( m \) is the mass, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change.
02

Calculate the Heat Lost by Silver

The specific heat capacity of silver is approximately 0.235 J/g°C. The temperature change \( \Delta T \) for silver is from 750.0°C to 0.0°C, which is \( 750.0 - 0.0 = 750.0 \)°C.Convert the mass of silver from kilograms to grams: 4.00 kg = 4000 g.Now, substitute into the formula: \[ q = 4000 \cdot 0.235 \cdot 750.0 \]Calculate \( q \).
03

Solve for Heat Energy

Now, let's perform the calculations:\[ q = 4000 \cdot 0.235 \cdot 750.0 = 705,000 \text{ J} \]So, the heat lost by the silver is 705,000 joules.
04

Calculate Mass of Ice Melted

To find out how much ice is melted, use the latent heat of fusion for ice which is 334 J/g. The formula is:\[ m_{\text{ice}} = \frac{q}{L_f} \]where \( L_f \) is the latent heat of fusion. Substituting the values:\[ m_{\text{ice}} = \frac{705000}{334} \]Calculate \( m_{\text{ice}} \).
05

Solve for Mass of Ice

Complete the calculation:\[ m_{\text{ice}} = \frac{705000}{334} \approx 2111 \text{ g} \]So, approximately 2111 grams of ice is melted.

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Key Concepts

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

Specific Heat Capacity
Every substance has a specific heat capacity, which is essentially the amount of heat required to raise the temperature of 1 gram of the substance by 1°C. For silver, this specific heat capacity is approximately 0.235 J/g°C. This means that it takes 0.235 joules of energy to increase the temperature of 1 gram of silver by one degree Celsius. When an object, like the silver ingot from our exercise, is placed in a new environment, its heat capacity determines how much energy it can transfer to or absorb from that environment.
This concept is central to calculating how much heat an object can release. The formula used for calculating the heat ( q ) in this scenario is:
  • q = m \cdot c \cdot \Delta T
where \(m\) is mass, \(c\) is specific heat capacity, and \(\Delta T\) is the change in temperature. Understanding how specific heat capacity works is crucial for solving problems involving temperature changes and energy transfer.
Latent Heat of Fusion
While specific heat capacity measures energy needed to alter temperature, latent heat of fusion describes the energy required to change a substance's state from solid to liquid without altering its temperature. For ice, this value is 334 J/g. This is the amount of energy needed to melt 1 gram of ice at its melting point, 0°C, into water at 0°C.
In our exercise, the energy released by the silver ingot as it cools is used to melt the ice. To calculate how much ice melts, we use the formula:
  • m_{\text{ice}} = \frac{q}{L_f}
where \(q\) is the heat energy absorbed by the ice and \(L_f\) is the latent heat of fusion. This formula helps us determine that approximately 2111 grams of ice was melted by the ingot.
Energy Conversion
Energy conversion is the process of changing energy from one form to another. In physical systems, energy is constantly being transformed; for example, in our exercise, thermal energy from the silver is converted into the energy required to melt ice. This is an example of heat energy being utilized in a thermodynamic process.
Understanding energy conversion is crucial in many scientific and engineering calculations. It highlights the idea that energy is conserved and can only change forms, not be created or destroyed. When the hot silver ingot cools down, its stored thermal energy decreases and becomes useful energy to change the phase of the ice from solid to liquid.
Physical Chemistry
This branch of chemistry explores how matter behaves on a molecular and atomic level and explains the physical properties of matter. It uses principles of physics and chemistry to study how systems and processes behave.
The concepts in this exercise, like heat transfer and phase changes, are deeply rooted in physical chemistry. When you heat or cool a substance, you're observing physical reactions that are driven by these chemical physics concepts. Understanding how heat affects a substance, such as triggering a phase change or causing it to absorb or release energy, is fundamental to this field. The exercise showcases these principles by displaying how heat from the silver ingot leads to the melting of ice, which involves understanding not just the chemical makeup, but the physical properties too.
Thermodynamics
Thermodynamics is the study of heat, energy, and work. It examines how energy is transferred in physical processes and how it affects matter. This science is crucial for understanding why and how heat transfer occurs, such as in the example with our silver ingot and ice.
The first law of thermodynamics, which states that energy cannot be created or destroyed, only transformed, is evident here. The system, in this case the silver ingot and the ice, must obey this principle. The energy from the silver does not disappear; instead, it's transferred to the ice to facilitate melting. Understanding these laws of energy distribution and conservation are vital in predicting the outcomes of thermal exchanges. Thermodynamics helps us comprehend not just this exercise, but broader phenomena in the natural world.

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Most popular questions from this chapter

(a) A typical student listening attentively to a physics lecture has a heat output of 100 W. How much heat energy does a class of 140 physics students release into a lecture hall over the course of a 50-min lecture? (b) Assume that all the heat energy in part (a) is transferred to the 3200 m\(^3\) of air in the room. The air has specific heat 1020 J/kg \(\cdot\) K and density 1.20 kg/m\(^3\). If none of the heat escapes and the air conditioning system is off, how much will the temperature of the air in the room rise during the 50-min lecture? (c) If the class is taking an exam, the heat output per student rises to 280 W. What is the temperature rise during 50 min in this case?

The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume e = 1 for these surfaces. Find the radii of the following stars (assumed to be spherical): (a) Rigel, the bright blue star in the constellation Orion, which radiates energy at a rate of \(2.7 \times 10{^3}{^2} W\) and has surface temperature 11,000 K; (b) Procyon B (visible only using a telescope), which radiates energy at a rate of \(2.1 \times 10{^2}{^3} W\) and has surface temperature 10,000 K. (c) Compare your answers to the radius of the earth, the radius of the sun, and the distance between the earth and the sun. (Rigel is an example of a supergiant star, and Procyon B is an example of a white dwarf star.)

A 500.0-g chunk of an unknown metal, which has been in boiling water for several minutes, is quickly dropped into an insulating Styrofoam beaker containing 1.00 kg of water at room temperature (20.0\(^\circ\)C). After waiting and gently stirring for 5.00 minutes, you observe that the water’s temperature has reached a constant value of 22.0\(^\circ\)C. (a) Assuming that the Styrofoam absorbs a negligibly small amount of heat and that no heat was lost to the surroundings, what is the specific heat of the metal? (b) Which is more useful for storing thermal energy: this metal or an equal weight of water? Explain. (c) If the heat absorbed by the Styrofoam actually is not negligible, how would the specific heat you calculated in part (a) be in error? Would it be too large, too small, or still correct? Explain.

In an effort to stay awake for an all-night study session, a student makes a cup of coffee by first placing a 200-W electric immersion heater in 0.320 kg of water. (a) How much heat must be added to the water to raise its temperature from 20.0\(^\circ\)C to 80.0\(^\circ\)C? (b) How much time is required? Assume that all of the heater’s power goes into heating the water.

What is the rate of energy radiation per unit area of a blackbody at (a) 273 K and (b) 2730 K?

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