Question

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

Approximately 0.32 kg of grape-flavored drink is needed.

Step-by-step solution

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°C.

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 \(Q_{ice} = 0.10 \cdot 334,000 = 33,400 \text{ J}\).

Calculating Heat Lost by Drink

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

Equating Heat Lost and Gained

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

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 \(m_{drink} \approx 0.32 \text{ kg}\).

Conclusion

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

Key concepts

Conservation of Energy

The Conservation of Energy is a foundational principle in physics that states that energy cannot be created or destroyed—only 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’s 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 \(4,186 \, \text{J/kg°C}\),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.