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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 \(28.0^{\circ} \mathrm{C},\) 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 that was initially at 80.0°C is 0.491 kg

Step by step solution

01

Calculate the heat required to raise the temperature of ice from -40°C to 0°C

The specific heat capacity of ice is \(c_{ice}= 2100 \, J/kg \cdot °C\). Apply the formula \(Q = mc\Delta T\), where \(\Delta T = T_{final} -T_{initial} = 0°C - (-40°C) = 40°C\), yielding \(Q_{iceto0} = 0.200 \, kg \times 2100 \, J/kg \cdot °C \times 40°C = 16800 \, J\)
02

Calculate the heat required to melt the ice from 0°C to water at 0°C

The heat of fusion for ice is \(L_f = 334000 \, J/kg\). We use the formula \(Q = mL_f\), yielding \(Q_{melt} = 0.200 \, kg \times 334000 \, J/kg = 66800 \, J\)
03

Calculate the heat required to raise the temperature of the water from ice from 0°C to 28°C

The specific heat capacity of water is \(c_{water} = 4186 \, J/kg \cdot °C\). Apply the formula \(Q = mc\Delta T\), with \(\Delta T = 28°C - 0°C = 28°C\), yielding \(Q_{0to28} = 0.200 \, kg \times 4186 \, J/kg \cdot °C \times 28°C = 23496 \, J\)
04

Calculate the total heat gained by the ice

The total heat gained by the ice (and subsequently water from the ice) is the sum of the heats calculated in the previous steps, \(Q_{total} = Q_{iceto0} + Q_{melt} + Q_{0to28} = 16800 \, J + 66800 \, J + 23496 \, J = 107096 \, J\)
05

Let's now talk about the water that was initially at 80.0°C

The water will lose heat, caused by the temperature going down to the final temperature of 28°C. Using the formula \(Q = mc\Delta T\), where \(m = m\) (the mass we are trying to find), \(c = 4186 \, J/kg \cdot °C\) (specific heat capacity of water) and \(\Delta T = T_{initial} - T_{final} = 80°C - 28°C = 52°C\). Set this equal to the total heat that was gained by the ice, which gives the equation: \(107096 \, J = m \times 4186 \, J/kg \cdot °C \times 52°C\)
06

Solve for m

Divide both sides of the equation by \(4186 \, J/kg \cdot °C \times 52°C\) to solve for the mass of the water: \(m = \frac{107096 \, J}{4186 \, J/kg \cdot °C \times 52°C} = 0.491 \, kg\)

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Key Concepts

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

Specific Heat Capacity
Understanding specific heat capacity is crucial when solving problems related to heat transfer, as seen in the textbook exercise. It represents the amount of energy required to raise the temperature of one kilogram of a substance by one degree Celsius (or one Kelvin).

This value is not only material-dependent but also crucial in determining how different substances react to heat input. For example, water has a relatively high specific heat capacity, meaning it requires more energy to change its temperature compared to other substances like metal. This is why water is such an effective coolant.

In our exercise, the specific heat capacities for ice and water are given, and we use them in the formula Q = mcΔT, where Q represents the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. A solid understanding of this property allows us to predict how much energy will be needed to change the temperature of a substance, which is essential for a wide range of applications, from climate science to culinary arts.
Heat of Fusion
When dealing with phase changes (such as the melting of ice to liquid water), the heat of fusion comes into play. Heat of fusion is the energy needed to change a substance from a solid to a liquid state at its melting point without changing its temperature.

This latent heat is characteristic of each material and, just like specific heat capacity, it tells us a lot about the substance's thermal properties. For ice to water transition, the heat of fusion is significant - 334,000 Joules per kilogram. This means quite a bit of energy is required to overcome the molecular forces holding the ice's crystalline structure together, which is provided in the form of heat during the melting process.

The formula used in the exercise, Q = mLf, where Q is the heat added or removed, m is the mass, and Lf is the heat of fusion, allows us to determine how much energy is necessary for the ice to completely melt into water. This knowledge is crucial not just in classroom exercises but in practical applications like designing refrigeration systems or understanding natural processes like glacier melting.
Thermal Equilibrium
The state of thermal equilibrium is a key concept to grasp in heat transfer scenarios. It occurs when two or more objects or substances in thermal contact reach the same temperature and exchange no more heat energy. At this point, the system is in balance, and there is a uniform temperature throughout.

In our textbook exercise, we leverage this idea to find the mass of water that, when mixed with ice, will result in a stable end temperature – in this case, 28°C. The energy given off by the warmer water as it cools is absorbed by the ice and vice versa until both substances reach the same final temperature and no more energy is exchanged. The calculations take us through the necessary steps to ensure that energy conservation is respected and thermal equilibrium is achieved.

Understanding thermal equilibrium is not only important for solving classroom problems but also for real-world applications, such as designing HVAC systems or creating comfortable living environments without wasting energy. These principles are at the very heart of thermodynamics.

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Most popular questions from this chapter

You have probably seen people jogging in extremely hot weather. There are good reasons not to do this! When jogging strenuously, an average runner of mass \(68 \mathrm{~kg}\) and surface area \(1.85 \mathrm{~m}^{2}\) produces energy at a rate of up to \(1300 \mathrm{~W}, 80 \%\) of which is converted to heat. The jogger radiates heat but actually absorbs more from the hot air than he radiates away. At such high levels of activity, the skin's temperature can be elevated to around \(33^{\circ} \mathrm{C}\) instead of the usual \(30^{\circ} \mathrm{C}\). (Ignore conduction, which would bring even more heat into his body.) The only way for the body to get rid of this extra heat is by evaporating water (sweating). (a) How much heat per second is produced just by the act of jogging? (b) How much net heat per second does the runner gain just from radiation if the air temperature is \(40.0^{\circ} \mathrm{C}\left(104^{\circ} \mathrm{F}\right) ?\) (Remember: He radiates out, but the environment radiates back in.) (c) What is the total amount of excess heat this runner's body must get rid of per second? (d) How much water must his body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is \(2.42 \times 10^{6} \mathrm{~J} / \mathrm{kg} .\) (e) How many \(750 \mathrm{~mL}\) bottles of water must he drink after (or preferably before!) jogging for a half hour? Recall that a liter of water has a mass of \(1.0 \mathrm{~kg}\).

You put a bottle of soft drink in a refrigerator and leave it until its temperature has dropped \(10.0 \mathrm{~K}\). What is its temperature change in (a) \(\mathrm{F}^{\circ}\) and \((\mathrm{b}) \mathrm{C}^{\circ} ?\)

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 \(70.0 \mathrm{~kg}\) 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 \(37.0^{\circ} \mathrm{C}\).

A \(\mathrm 500.0 \mathrm{~g}\) 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) 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.

Animals in cold climates often depend on \(t w o\) layers of insulation: a layer of body fat (of thermal conductivity \(0.20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\) ) surrounded by a layer of air trapped inside fur or down. We can model a black bear (Ursus americanus) as a sphere \(1.5 \mathrm{~m}\) in diameter having a layer of fat \(4.0 \mathrm{~cm}\) thick. (Actually, the thickness varies with the season, but we are interested in hibernation, when the fat layer is thickest.) In studies of bear hibernation, it was found that the outer surface layer of the fur is at \(2.7^{\circ} \mathrm{C}\) during hibernation, while the inner surface of the fat layer is at \(31.0^{\circ} \mathrm{C}\). (a) What is the temperature at the fat-inner fur boundary so that the bear loses heat at a rate of \(50.0 \mathrm{~W} ?\) (b) How thick should the air layer (contained within the fur) be?

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