Question

A \(0.3-\mathrm{L}\) glass of water at \(20^{\circ} \mathrm{C}\) is to be cooled with ice to \(5^{\circ} \mathrm{C}\). Determine how much ice needs to be added to the water, in grams, if the ice is at \((a) 0^{\circ} \mathrm{C}\) and \((b)-20^{\circ} \mathrm{C}\) Also determine how much water would be needed if the cooling is to be done with cold water at \(0^{\circ} \mathrm{C}\). The melting temperature and the heat of fusion of ice at atmospheric pressure are \(0^{\circ} \mathrm{C}\) and \(333.7 \mathrm{kJ} / \mathrm{kg}\), respectively, and the density of water is \(1 \mathrm{kg} / \mathrm{L}\).

Short answer

To cool down 0.3 L of water from 20°C to 5°C, the following amounts of ice/cold water are required: 1. For ice at 0°C: approximately 56 g of ice. 2. For ice at -20°C: approximately 50 g of ice. Alternatively, if using cold water at 0°C, approximately 180 g of cold water would be needed.

Step-by-step solution

List the given values

We are given the following values: - Volume of water (\(V_{water}) = 0.3 L\) - Initial temperature of water (\(T_{initial}) = 20^{\circ} C\) - Desired temperature of water (\(T_{final}) = 5^{\circ} C\) - Ice temperatures: \((a) 0^{\circ} C\) and \((b) -20^{\circ} C\) - Melting temperature of ice (\(T_{melt}) = 0^{\circ} C\) - Heat of fusion of ice (\(L_f) = 333.7 kJ/kg\) - Density of water (\(\rho_{water}) = 1 kg/L\) - Specific heat capacity of water (\(c_{water}) = 4.18 kJ/(kg \cdot K)\) - Specific heat capacity of ice (\(c_{ice}) = 2.1 kJ/(kg \cdot K)\)

Calculate the mass of the water

Since we will deal with the heat transfer and using energy balance equation, we must calculate the mass of the water and convert it to kilograms. \(mass_{water} = V_{water} \times \rho_{water} = 0.3 L \times 1 kg / L = 0.3 kg\)

Determine the energy balance equations

To cool down the water with ice, heat must be transferred from the water to the ice. We need to make sure all the energy is balanced in the system. The energy balance equation depends on the initial temperature of the ice. Let \(mass_{ice}\) represent the mass of ice needed. 1. For ice at \(0^{\circ} C\), the ice will only absorb energy to melt, so the energy balance equation is: \(Q_{water\_to\_cool} = Q_{ice\_to\_melt}\) 2. For ice at \(-20^{\circ} C\), the energy balance equation has two parts: energy required to heat up the ice to \(0^{\circ} C\), and energy required to melt the ice: \(Q_{water\_to\_cool} = Q_{ice\_to\_heat} + Q_{ice\_to\_melt}\)

Calculate the heat transfer for each scenario

1. For ice at \(0^{\circ} C\): We must calculate the heat transfer from water to cool down to the required final temperature: \(Q_{water\_to\_cool} = mass_{water} \times c_{water} \times (T_{initial} - T_{final}) = 0.3 kg \times 4.18 kJ/(kg \cdot K) \times (20 - 5) = 18.81 kJ\) Now, calculate the mass of ice needed to absorb that energy to melt: \(Q_{ice\_to\_melt} = mass_{ice} \times L_f\) Solving for \(mass_{ice}\): \(mass_{ice} = \dfrac{Q_{water\_to\_cool}}{L_f} = \dfrac{18.81 kJ}{333.7 kJ/kg} = 0.056 kg\) Therefore, \(mass_{ice}\approx 56 g\) for ice at \(0^{\circ} C\). 2. For ice at \(-20^{\circ} C\): First, calculate the heat transfer to heat the ice to \(0^{\circ} C\) before melting: \(Q_{ice\_to\_heat} = mass_{ice} \times c_{ice} \times (T_{melt} - T_{initial})\) Now, set up the energy balance equation: \(Q_{water\_to\_cool} = Q_{ice\_to\_heat} + Q_{ice\_to\_melt}\) Substituting the values: \(18.81 kJ = mass_{ice} \times (2.1 kJ/(kg \cdot K) \times (0 - (-20)) + 333.7 kJ/kg)\) Solving for \(mass_{ice}\), we get: \(mass_{ice} = \dfrac{18.81 kJ}{(2.1 kJ/(kg \cdot K) \times 20 + 333.7 kJ/kg)} = \dfrac{18.81 kJ}{376.7 kJ/kg} = 0.050 kg\) Therefore, \(mass_{ice}\approx 50 g\) for ice at \(-20^{\circ} C\).

Calculate the amount of cold water needed

Finally, we need to find out how much cold water at \(0^{\circ} C\) would be needed instead of ice. Let's call the mass of cold water needed \(mass_{cold\_water}\). The energy balance equation will be: \(Q_{water\_to\_cool} = Q_{cold\_water\_to\_heat}\) Now, calculate the heat transfer from the cold water to heat up to the final temperature: \(Q_{cold\_water\_to\_heat} = mass_{cold\_water} \times c_{water} \times (T_{final} - T_{melt})\) Solve for \(mass_{cold\_water}\): \(mass_{cold\_water} = \dfrac{Q_{water\_to\_cool}}{c_{water} \times (T_{final} - T_{melt})} = \dfrac{18.81 kJ}{4.18 kJ/(kg \cdot K) \times (5 - 0)} = 0.180 kg\) Therefore, \(mass_{cold\_water} \approx 180 g\) of cold water at \(0^{\circ} C\) would be required.

Key concepts

Heat Transfer

Understanding heat transfer is key to solving problems involving changes in temperature or state of a substance. Heat transfer, simply put, is the movement of thermal energy from one object to another as a result of a temperature difference. In the context of our exercise, when we add ice to water, heat transfer occurs from the warmer water to the colder ice. This process will continue until thermal equilibrium is reached, meaning the water and ice eventually settle at a common temperature.

The rate of heat transfer is influenced by several factors, including the temperature difference between the substances, the thermal conductivity of the materials, and the surface area over which the heat is transferred. In the case of cooling water with ice, two key physical factors come into play: the specific heat capacity of water and the heat of fusion of ice. The specific heat capacity governs how much energy is required to change the temperature of the water, and the heat of fusion describes the energy needed for the ice to transition from a solid to a liquid state at its melting point.

Energy Balance

Energy balance is a fundamental concept in thermodynamics which states that energy within a closed system must be conserved. It is often visualized as an accounting of energy inputs, outputs, and storage changes within the system. In our glass of water and ice scenario, the principle of energy balance is used to determine the correct amount of ice that needs to be added to achieve the desired water temperature.

The process involves equating the heat lost by the water, as it cools, to the heat gained by the ice, as it melts or warms up to the melting point before melting. To cool the water down from 20°C to 5°C, we calculate the heat loss using the mass of the water and its specific heat capacity. Then, for the energy balance, the heat lost by the water must equal the heat gained by the ice, which can be either the heat absorbed while the ice is melting at 0°C or, when at -20°C, the sum of the heat absorbed while warming to 0°C and then melting.

Specific Heat Capacity

The specific heat capacity is a property of a material that indicates how much heat energy is required to raise the temperature of one kilogram of the substance by one degree Celsius (or one Kelvin). It is an intrinsic property that tells us a great deal about how substances respond to the addition or removal of heat. For example, water has a high specific heat capacity, which means it takes a considerable amount of heat to raise its temperature compared to other substances.

In the exercise, we use the specific heat capacity of water (4.18 kJ/kg·K) to find out how much energy is needed to cool down from 20°C to 5°C. The specific heat capacity plays a major role in our calculations for determining the mass of ice needed for cooling. It also comes into play when we calculate the alternative scenario of using cold water instead of ice to achieve the same cooling effect. This underscores the specific heat capacity's importance in understanding heat transfer and energy balance in thermal systems.