Question

A glass of water with a mass of \(0.45 \mathrm{kg}\) at \(20^{\circ} \mathrm{C}\) is to be cooled to \(0^{\circ} \mathrm{C}\) by dropping ice cubes at \(0^{\circ} \mathrm{C}\) into it. The latent heat of fusion of ice is \(334 \mathrm{kJ} / \mathrm{kg}\), and the specific heat of water is \(4.18 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C} .\) The amount of ice that needs to be added is \((a) 56 \mathrm{g}\) (b) \(113 \mathrm{g}\) \((c) 124 \mathrm{g}\) \((d) 224 \mathrm{g}\) \((e) 450 \mathrm{g}\)

Short answer

Answer: \(113 \, \mathrm{g}\)

Step-by-step solution

Calculate the heat removed from the water to reach \(0^{\circ} \mathrm{C}\)

To calculate the heat removed from the water (\(Q_{water}\)), we can use the formula \(Q = mcΔT\), where \(m\) is the mass of the water, \(c\) is the specific heat of water, and \(ΔT\) is the temperature change of the water. In this case, \(m = 0.45 \, \mathrm{kg}\), \(c = 4.18 \, \mathrm{kJ/kg\cdot^{\circ} C}\), and \(ΔT = 20 - 0 = 20\, ^{\circ} \mathrm{C}\). \(Q_{water} = (0.45 \, \mathrm{kg}) \times (4.18 \, \mathrm{kJ/kg\cdot^{\circ} C}) \times (20 \,^{\circ} \mathrm{C}) = 37.62 \, \mathrm{kJ}\)

Calculate the heat absorbed by the ice per unit mass while melting

To calculate the heat absorbed by the ice (\(Q_{ice}\)) per unit mass while melting, we can use the latent heat of fusion of ice \(L = 334 \, \mathrm{kJ/kg}\). \(Q_{ice\_per\_unit\_mass} = 334 \, \mathrm{kJ/kg}\)

Determine the mass of ice needed

To find the mass of ice (\(m_{ice}\)) needed, we can equate the heat removed from the water to the heat absorbed by the ice while melting \(Q_{water} = m_{ice} \times Q_{ice\_per\_unit\_mass}\). Thus, \(m_{ice} = \frac{Q_{water}}{Q_{ice\_per\_unit\_mass}} = \frac{37.62 \, \mathrm{kJ}}{334 \, \mathrm{kJ/kg}} = 0.1126 \, \mathrm{kg}\) To convert this value to grams, we can multiply it by \(1000\): \(0.1126 \, \mathrm{kg} \times 1000 = 112.6 \, \mathrm{g}\), which is approximately equal to \(113 \, \mathrm{g}\). Therefore, the correct answer is \((b) \, 113 \, \mathrm{g}\).

Key concepts

Latent Heat of Fusion

The term latent heat of fusion refers to the amount of heat energy required to change a substance from the solid phase to the liquid phase without changing its temperature. This process is also known as melting. For example, ice melting into water requires absorption of energy; likewise, when water freezes, it releases energy. The amount of heat needed is typically expressed in terms of energy per unit mass, often in joules per kilogram (J/kg).

In the given exercise, the latent heat of fusion of ice is stated as 334 kJ/kg. This high energy requirement explains why ice at 0°C doesn't instantly turn to water, even when it's at the melting point. It needs substantial energy to overcome the forces holding its molecules in a solid structure.

Understanding this concept is crucial when calculating the amount of ice needed to cool water, as in our example. Only by melting and absorbing the correct amount of heat can the ice achieve the desired cooling effect.

Specific Heat Capacity

The specific heat capacity of a substance is a measure of the amount of heat energy required to raise the temperature of one kilogram of the substance by one degree Celsius (or one Kelvin). The unit of measurement is typically joules per kilogram per degree Celsius (J/kg°C).

Water, for instance, has a specific heat capacity of 4.18 kJ/kg°C, which is relatively high. This means that it can absorb a lot of heat before its temperature rises significantly. This property is why water is an effective coolant and is also why changes in water temperature play a significant role in Earth's climate.

When solving thermodynamic problems, knowing the specific heat capacity enables us to calculate the energy change associated with temperature changes. This is essential for our exercise as we initially determine how much energy is removed from the water as it is cooled from 20°C to 0°C.

Temperature Change in Thermodynamics

In thermodynamics, temperature change represents the variation in the thermal state of a substance when heat is added or removed. When a substance heats up, its temperature rises, while removing heat causes its temperature to drop. Understanding temperature change is vital for solving problems involving heat transfer.

In the provided exercise, the water's initial temperature is 20°C, and the goal is to cool it to 0°C. The temperature change (ΔT), in this case, is a reduction of 20°C. Utilizing the specific heat capacity of water, we can calculate the exact amount of heat energy taken out of the water to achieve this cooling effect.

Recognizing how temperature affects the state of a substance – whether it remains in the same state, transitions to a solid (freezes), or transitions to a gas (evaporates) – is fundamental in thermodynamics. Therefore, analyzing temperature changes is key to managing energy transfers and achieving desired outcomes in various processes, including the melting of ice to cool a glass of water.