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Chapter 14: Problem 100

On a very hot summer day, Daphne is off to the park for a picnic. She puts \(0.10 \mathrm{kg}\) of ice at \(0^{\circ} \mathrm{C}\) in a thermos and then adds a grape-flavored drink, which she has mixed from a powder using room temperature water \(\left(25^{\circ} \mathrm{C}\right) .\) How much grape- flavored drink will just melt all the ice?

Short Answer

Expert verified
Approximately 0.32 kg of grape-flavored drink is needed.

Step by step solution

01

Understanding Heat Transfer

In this problem, the heat lost by the grape-flavored drink will equal the heat gained by the ice to melt. We will use the principle of conservation of energy. The energy required to melt the ice will be equal to the energy lost by the drink as it cools to 0掳C.
02

Calculating Heat Required to Melt Ice

The heat required to melt ice can be calculated using the formula \(Q_{ice} = m_{ice} \cdot L_f\), where \(m_{ice} = 0.10 \text{ kg}\) is the mass of the ice and \(L_f = 334,000 \text{ J/kg}\) is the latent heat of fusion for ice. Calculating this gives \(Q_{ice} = 0.10 \cdot 334,000 = 33,400 \text{ J}\).
03

Calculating Heat Lost by Drink

The heat lost by the grape drink when it cools to 0掳C is given by \(Q_{drink} = m_{drink} \cdot c_{water} \cdot \Delta T\), where \(m_{drink}\) is the mass of the drink, \(c_{water} = 4,186 \text{ J/kg掳C}\) is the specific heat capacity of water, and \(\Delta T = 25 - 0 = 25掳C\) is the temperature change. The equation becomes \(Q_{drink} = m_{drink} \cdot 4,186 \cdot 25\).
04

Equating Heat Lost and Gained

Set the heat lost by the drink equal to the heat gained by the ice: \(m_{drink} \cdot 4,186 \cdot 25 = 33,400.\)
05

Solving for Mass of Drink

Rearrange the equation to solve for \(m_{drink}\): \(m_{drink} = \frac{33,400}{4,186 \cdot 25}\). Calculating this gives \(m_{drink} \approx 0.32 \text{ kg}\).
06

Conclusion

Therefore,\ the mass of the grape-flavored drink needed to just melt all the ice is approximately \(0.32 ext{ kg}\).

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Key Concepts

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

Conservation of Energy
The Conservation of Energy is a foundational principle in physics that states that energy cannot be created or destroyed鈥攐nly transformed or transferred between systems. In simple terms, the total energy in a closed system remains constant.
  • In our exercise, the grape-flavored drink and ice system are treated as closed, meaning the heat energy lost by the drink equals the energy needed to melt the ice.
  • It鈥檚 essential to understand that as the grape drink cools down, its thermal energy is transferred to the ice as heat, leading the ice to melt.
Recognizing this concept in real-world applications, like melting ice with a warmer liquid, helps in calculating necessary thermal exchanges, as we did when we equated the energy transferred from the drink to ice.
Latent Heat of Fusion
Latent Heat of Fusion refers to the amount of energy needed to change a substance's state from solid to liquid without changing its temperature. For ice, the latent heat of fusion is significant because it quantifies how much energy is required to melt it.
  • The formula for calculating the energy required to melt ice is: \[Q_{ice} = m_{ice} \cdot L_f\]where \(L_f = 334,000 \, \text{J/kg}\).
  • In our situation, with \(m_{ice} = 0.10 \, \text{kg}\),the energy required is \(33,400 \, \text{J}\). This means a significant amount of thermal energy must be absorbed for the ice to melt, even without a temperature increase.
This concept is especially important for understanding energy exchanges involving phase changes, where temperature remains constant but internal energy changes significantly.
Specific Heat Capacity
Specific Heat Capacity is a property that describes how much heat is needed to change the temperature of a unit mass of a substance by one degree Celsius. Water, for example, has a high specific heat capacity at \(4,186 \, \text{J/kg掳C}\),making it efficient at storing heat energy.
  • The specific heat capacity of water plays a crucial role in our exercise by determining how much the water (drink) must change in order to provide the required energy to melt the ice.
  • In our case, the specific heat equation is:\[Q_{drink} = m_{drink} \cdot c_{water} \cdot \Delta T\]helping us calculate how much energy the drink can lose when cooled.
Understanding specific heat helps predict how substances respond to heating and cooling in various practical scenarios, such as cooking or meteorological phenomena.
Temperature Change
Temperature Change indicates how much the temperature of a substance varies, often due to the addition or removal of heat energy. In our exercise, the temperature change helps in estimating the energy transitions without considering any phase changes.
  • We calculate temperature change using\(\Delta T = T_{initial} - T_{final}\).For our grape-flavored drink, initially at \(25^{\circ} \text{C}\) and cooling to \(0^{\circ} \text{C}\), \(\Delta T = 25\) degrees.
  • The temperature change directly affects how much heat is extracted from the drink, allowing it to melt the ice.
Understanding how temperature change influences heat flow is crucial in fields ranging from domestic heating and cooling systems to industrial processes and climate science.

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Most popular questions from this chapter

If a leaf is to maintain a temperature of \(40^{\circ} \mathrm{C}\) (reasonable for a leaf), it must lose \(250 \mathrm{W} / \mathrm{m}^{2}\) by transpiration (evaporative heat loss). Note that the leaf also loses heat by radiation, but we will neglect this. How much water is lost after 1 h through transpiration only? The area of the leaf is \(0.005 \mathrm{m}^{2}\).
A \(0.500-\mathrm{kg}\) slab of granite is heated so that its temperature increases by \(7.40^{\circ} \mathrm{C} .\) The amount of heat supplied to the granite is \(2.93 \mathrm{kJ} .\) Based on this information, what is the specific heat of granite?

A copper rod of length \(0.50 \mathrm{m}\) and cross-sectional area \(6.0 \times 10^{-2} \mathrm{cm}^{2}\) is connected to an iron rod with the same cross section and length \(0.25 \mathrm{m} .\) One end of the copper is immersed in boiling water and the other end is at the junction with the iron. If the far end of the iron rod is in an ice bath at \(0^{\circ} \mathrm{C},\) find the rate of heat transfer passing from the boiling water to the ice bath. Assume there is no heat loss to the surrounding air. (tutorial: composite rod)
What mass of water at \(25.0^{\circ} \mathrm{C}\) added to a Styrofoam cup containing two 50.0 -g ice cubes from a freezer at \(-15.0^{\circ} \mathrm{C}\) will result in a final temperature of \(5.0^{\circ} \mathrm{C}\) for the drink?
A hiker is wearing wool clothing of \(0.50-\mathrm{cm}\) thickness to keep warm. Her skin temperature is \(35^{\circ} \mathrm{C}\) and the outside temperature is \(4.0^{\circ} \mathrm{C} .\) Her body surface area is \(1.2 \mathrm{m}^{2} .\) (a) If the thermal conductivity of wool is $0.040 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K}),$ what is the rate of heat conduction through her clothing? (b) If the hiker is caught in a rainstorm, the thermal conductivity of the soaked wool increases to \(0.60 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})\) (that of water). Now what is the rate of heat conduction?
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