Question

\(500 \mathrm{~g}\) of ice at \(0^{\circ} \mathrm{C}\) is added to \(500 \mathrm{~g}\) of water at \(64^{\circ} \mathrm{C}\). When the temperature of the mixture is \(0^{\circ} \mathrm{C}\), what weight of ice is still present? Heat of fusion of \(\mathrm{H}_{2} \mathrm{O}=80 \mathrm{cal} / \mathrm{g}\).

Short answer

The weight of the remaining ice when the mixture reaches \(0^{\circ} \mathrm{C}\) is 600g.

Step-by-step solution

Determine the heat gained by the melting ice

To determine the heat gained by the initial ice as it melts, we use the heat of fusion of water, which is 80 cal/g. The heat gained by the ice (Q_ice) can be calculated using the formula: \(Q_{\text{ice}} = m_{\text{ice}} \times L\) where \(m_{\text{ice}}\) is the mass of the ice (500g), and \(L\) is the heat of fusion of water (80 cal/g). \(Q_{\text{ice}} = 500g \times 80\frac{cal}{g}\) \(Q_{\text{ice}} = 40,000 cal\)

Determine the heat lost by the water as it cools down

The specific heat capacity of water is 1 cal/(g°C). The heat lost by the water (Q_water) as it cools down can be calculated using the formula: \(Q_{\text{water}} = m_{\text{water}} \times c \times \Delta T\) where \(m_{\text{water}}\) is the mass of the water (500g), \(c\) is the specific heat capacity of water (1 cal/(g°C)), and \(\Delta T\) is the difference in temperature (64°C - 0°C). \(Q_{\text{water}} = 500g \times 1\frac{cal}{g^\circ C} \times 64^\circ C\) \(Q_{\text{water}} = 32,000 cal\)

Calculate the remaining heat available for melting the additional ice

We have calculated the heat gained by the initial ice and the heat lost by the water. The remaining heat available for melting the additional ice can be calculated by subtracting the heat gained by melting initial ice from the heat lost by water. \(Q_{\text{remaining}} = Q_{\text{water}} - Q_{\text{ice}}\) \(Q_{\text{remaining}} = 32,000 cal - 40,000 cal\) \(Q_{\text{remaining}} = -8,000 cal\) Since the remaining heat is a negative value, it means that not all the initial ice melts due to the inability of the water to provide enough heat.

Determine the amount of ice melted and the weight of the remaining ice

To determine the amount of ice melted, we will use the remaining heat available and the heat of fusion of water (80 cal/g). We can determine the mass of the melted ice (m_melted_ice) using the formula: \(m_{\text{melted\_ice}} = \frac{Q_{\text{remaining}}}{L}\) \(m_{\text{melted\_ice}} = \frac{-8,000 cal}{80 \frac{cal}{g}}\) \(m_{\text{melted\_ice}} = -100g\) Since the mass of the melted ice is a negative value, it means that 100g of the initial ice did not melt. Therefore, the weight of the remaining ice when the mixture reaches 0°C is: Weight of remaining ice = Initial mass of the ice - |Mass of melted ice| Weight of remaining ice = 500g - |-100g| Weight of remaining ice = 500g + 100g Weight of remaining ice = 600g

Key concepts

Heat of Fusion

The heat of fusion is an important principle in understanding phase changes, particularly the transition between solid and liquid states. When we talk about the heat of fusion, we're discussing the amount of heat energy needed to convert a solid into a liquid at a constant temperature. For water, this value is 80 calories per gram (cal/g).
In the original exercise, you are given an example where ice at 0°C is introduced into water. To dissolve the ice, heat from the warmer water is used to melt the ice. This process consumes 80 calories for each gram of ice and needs to be considered when solving thermodynamic problems involving phase change.
Remember, the process of melting doesn't increase the temperature of the ice itself; rather, it changes its state. This energy is essential for overcoming the bonds that hold the solid ice together.

Specific Heat Capacity

The specific heat capacity tells us how much heat energy is required to raise the temperature of a particular substance by a given amount, usually one degree Celsius. For water, the specific heat capacity is 1 cal/(g°C). This means that it requires a single calorie to increase the temperature of 1 gram of water by one degree Celsius.

• Specific heat capacity differs from heat of fusion as it relates to temperature change, not phase change.
• It plays a critical role in calculating the heat balance when different substances interact thermally.

This concept is crucial when calculating the heat lost by the water as it cools in our exercise. The specific heat capacity helps determine how much thermal energy is released when the water temperature drops, which in turn influences the melting of the ice.

Melting Point

The melting point is the temperature at which a substance transitions from solid to liquid. For ice, the melting point is at 0°C or 32°F under standard atmospheric pressure. At this precise temperature, ice starts to melt and turn into water without any temperature increase until it fully transforms.
This understanding is key in the given problem, as the setup involves ice and water interacting at exactly 0°C. Due to its melting point, ice used as a coolant forces us to consider both heat it can absorb as it melts and the heat required for the phase transformation.
Remember that even though substances have a specific melting point, the actual process of melting can vary based on the amount of heat available and conditions like pressure.

Thermodynamics

Thermodynamics is the science that deals with the relationships between heat and other forms of energy during various physical processes. It's a broad field that helps us understand how energy is transferred and conserved.
In the context of this exercise, thermodynamics plays a crucial role in understanding how heat is exchanged between the ice and water until thermal equilibrium is reached. The primary principles at play here include the conservation of energy and heat transfer.

• Energy conservation states that the total energy in a closed system remains constant.
• Heat transfer occurs from the hotter water to the cooler ice until the system reaches equilibrium.

In our example, the principles of thermodynamics help explain why not all the ice melts: the available heat from the water is insufficient to fully melt the ice, resulting in a final mixture temperature of 0°C.