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In a container of negligible mass, 0.200 \(\mathrm{kg}\) of ice at an initial temperature of \(-40.0^{\circ} \mathrm{C}\) is mixed with a mass \(m\) of water that has an initial temperature of \(80.0^{\circ} \mathrm{C}\) . No heat is lost to the surroundings. If the final temperature of the system is \(20.0^{\circ} \mathrm{C},\) what is the mass \(m\) of the water that was initially at \(80.0^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
The mass of the water initially at 80°C is 0.400 kg.

Step by step solution

01

Understanding the Heat Transfer

Since no heat is lost to the surroundings, the heat gained by the ice must be equal to the heat lost by the water. We calculate the heat required to raise ice from \(-40.0^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\), melt it, and then raise the resulting water to \(20.0^{\circ}\mathrm{C}\). We equate this to the heat lost by the water as it cools from \(80.0^{\circ} \mathrm{C}\) to \(20.0^{\circ}\).
02

Heat Required to Warm Ice to 0°C and Melt it

Calculate the heat to warm ice: \[ q_1 = m_{\text{ice}} \cdot c_{\text{ice}} \cdot \Delta T \]where \( m_{\text{ice}} = 0.200\, \mathrm{kg}, \ c_{\text{ice}} = 2.1\, \mathrm{J/g·^{\circ}C}, \ \Delta T = 40^{\circ} \mathrm{C}\).Convert mass to grams: \( 0.200 \times 1000 = 200 \mathrm{g}\).So, \[ q_1 = 200 \cdot 2.1 \cdot 40 = 16800 \, \mathrm{J} \]Then, calculate the heat to melt ice:\[ q_2 = m_{\text{ice}} \cdot L_f \]where \( L_f = 334\, \mathrm{J/g}\).\[ q_2 = 200 \cdot 334 = 66800 \, \mathrm{J} \]
03

Heat Required to Raise Melted Water to 20°C

Calculate the heat for raising the temperature of water from \(0^{\circ} \mathrm{C}\) (after melting) to \(20^{\circ} \mathrm{C}\):\[ q_3 = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T \]where \(c_{\text{water}} = 4.18\, \mathrm{J/g·^{\circ}C}\) and \(\Delta T = 20^{\circ} \mathrm{C}\).\[ q_3 = 200 \cdot 4.18 \cdot 20 = 16720 \, \mathrm{J} \]
04

Total Heat Absorbed by the Ice

Sum up all the heats calculated:\[ q_{\text{total}} = q_1 + q_2 + q_3 = 16800 + 66800 + 16720 = 100320 \, \mathrm{J} \]
05

Heat Lost by the Water Initially at 80°C

Let's calculate the heat lost by the warmer water:\[ q_{\text{water}} = m \cdot c_{\text{water}} \cdot \Delta T \]where \(\Delta T = 60^{\circ} \mathrm{C}\) (from \(80^{\circ} \mathrm{C}\) to \(20^{\circ} \mathrm{C}\)).\[ q_{\text{water}} = m \cdot 4.18 \cdot 60 = 250.8m \, \mathrm{J} \]
06

Set the Heat Gained by Ice Equal to Heat Lost by Water

Since all the heat gained by the ice comes from the water:\[ q_{\text{total}} = q_{\text{water}} \]\[ 100320 = 250.8m \]Solve for \( m \):\[ m = \frac{100320}{250.8} = 400.0 \, \mathrm{g} \]Convert to kg: \[ 400 \, \mathrm{g} = 0.400 \, \mathrm{kg} \].

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Key Concepts

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

Thermodynamics
Thermodynamics is the study of energy and its transformations. It focuses on understanding how energy moves and changes within a given system. One important aspect of thermodynamics is heat transfer. Heat transfer refers to the movement of thermal energy from one object or substance to another, based on temperature differences.
In the given exercise, heat transfer plays a central role. The heat lost by the warmer water is transferred to the ice, causing it to eventually reach an equilibrium temperature with the water at 20°C. This performance demonstrates the First Law of Thermodynamics, also known as the principle of energy conservation. It states that, within an isolated system, energy can neither be created nor destroyed; it can only be transferred or transformed.
Every problem involving thermodynamics requires analyzing the transfer of energy between substances and ensuring that this energy conforms to the laws of thermodynamics.
Specific Heat Capacity
Specific heat capacity refers to the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius. Different materials absorb and release heat at different rates, and their specific heat capacities greatly influence these rates.
In this exercise, we used specific heat capacities for ice and water to calculate the thermal energy changes. The specific heat capacity of ice is lower than that of water, which means it requires less energy to raise the temperature of ice compared to an equivalent mass of water over the same temperature range:
  • Specific heat capacity of ice: 2.1 J/g°C
  • Specific heat capacity of water: 4.18 J/g°C
The high heat capacity of water allows it to store considerable energy, explaining why water is effective in regulating Earth's climate and providing natural thermal stability.
Phase Change
Phase change refers to the transition of a substance from one state of matter to another, such as from solid to liquid or liquid to gas. These transitions require or release heat without changing the temperature of the substance during the process.
In the exercise, we observed a phase change as ice melted to become water. The energy required for this phase change is known as the latent heat of fusion. This value represents the amount of heat needed to convert ice at 0°C to water at the same temperature:
  • Latent heat of fusion ( L_f ): 334 J/g
During phase change, energy goes into breaking or forming intermolecular bonds, rather than increasing the substance's kinetic energy, which is why the temperature remains constant. Understanding phase changes is critical in thermodynamics, as they occur widely in nature and technological processes.
Energy Conservation
Energy conservation is a core concept in physics, which states that the total energy of an isolated system remains constant, although energy can transform from one form to another. This principle underpins the First Law of Thermodynamics.
In the problem, no heat is lost to the surroundings, highlighting energy conservation. The energy absorbed by the ice and cold water exactly equals the energy lost by the hot water. This approach lets us set up the equation:\( q_{\text{total}} = q_{\text{water}} \)and find the mass of the warmer water.
Energy conservation principles are fundamental to solving any heat transfer problems in thermodynamics, as they ensure that all energy exchanges are accounted for in closed systems.

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Most popular questions from this chapter

A metal rod that is 30.0 \(\mathrm{cm}\) long expands by 0.0650 \(\mathrm{cm}\) when its temperature is raised from \(0.0^{\circ} \mathrm{C}\) to \(100.0^{\circ} \mathrm{C} .\) A rod of a different metal and of the same length expands by 0.0350 \(\mathrm{cm}\) for the same rise in temperature. A third rod, also 30.0 \(\mathrm{cm}\) long, is made up of pieces of each of the above metals placed end to end and expands 0.0580 \(\mathrm{cm}\) between \(0.0^{\circ} \mathrm{C}\) and \(100.0^{\circ} \mathrm{C} .\) Find the length of each portion of the composite rod.

\(\mathrm{A} 500.0-\mathrm{g}\) 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 \(\mathrm{kg}\) of water at room temperature \(\left(20.0^{\circ} \mathrm{C}\right) .\) After waiting and gently stirring for 5.00 minutes, you observe that the water's temperature has reached a constant value of \(22.0^{\circ} \mathrm{C}\) (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) What 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.

A person of mass 70.0 \(\mathrm{kg}\) is sitting in the bathtub. The bathtub is 190.0 \(\mathrm{cm}\) by 80.0 \(\mathrm{cm}\) ; before the person got in, the water was 16.0 \(\mathrm{cm}\) deep. The water is at a temperature of \(37.0^{\circ} \mathrm{C}\) . Suppose that the water were to cool down spontaneously to form ice at \(0.0^{\circ} \mathrm{C},\) and that all the energy released was used to launch the hapless bather vertically into the air. How high would the bather go? (As you will see in Chapter 20 , this event is allowed by energy conservation but is prohibited by the second law of thermodynamics.)

The emissivity of tungsten is \(0.350 .\) A tungsten sphere with radius 1.50 \(\mathrm{cm}\) is suspended within a large evacuated enclosure whose walls are at 290.0 \(\mathrm{K}\) . What power input is required to maintain the sphere at a temperature of 3000.0 \(\mathrm{K}\) if heat conduction along the supports is neglected?

Treatment for a Stroke. One suggested treatment for a person who has suffered a stroke is immersion in an ice-water bath at \(0^{\circ} \mathrm{C}\) to lower the body temperature, which prevents damage to the brain. In one set of tests, patients were cooled until their internal temperature reached \(32.0^{\circ} \mathrm{C}\) . To treat a 70.0 -kg patient, what is the minimum amount of ice (at \(0^{\circ} \mathrm{C}\) ) you need in the bath so that its temperature remains at \(0^{\circ} \mathrm{C} ?\) The specific heat of the human body is \(3480 \mathrm{J} / \mathrm{kg} \cdot \mathrm{C}^{\circ},\) and recall that normal body temperature is \(37.0^{\circ} \mathrm{C} .\)

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