Question

(a) Two \(50-\mathrm{g}\) ice cubes are dropped into \(200 \mathrm{~g}\) of water in a glass. If the water were initially at a temperature of \(25^{\circ} \mathrm{C}\), and if the ice came directly from a freezer at \(-15^{\circ} \mathrm{C}\), what is the final temperature of the drink? ( \(b\) ) If only one ice cube had been used in (a), what would be the final temperature of the drink? Neglect the heat capacity of the glass.

Short answer

For part (a), the final temperature of the drink is approximately 14.6°C. For part (b), it would be approximately 19.8°C.

Step-by-step solution

Calculate the Heat Required to Warm the Ice to 0 Degrees Celsius

The heat (Q) absorbed or dissolved by a body when it's heated or cooled can be calculated using the formula Q = mcΔT, where m is the mass of the substance (in this case, ice), c is the specific heat capacity of the substance and ΔT is the change in temperature. For Ice, c = 2.09 J/g°C. So, the heat required to warm the ice to 0°C can be calculated as Q1 = (50 g)*(2.09 J/g°C)*(-15°C-0°C).

Calculate the Heat Required to Melt Ice into Water at 0 Degrees Celsius

When a substance changes state, the heat involved (Q2) can be calculated using the formula Q = mL, where m is the mass and L is the latent heat. For ice, L = 334 J/g. The heat to turn this ice into water can therefore be calculated as Q2 = (50 g)*(334 J/g).

Find the Sum of Heats Dissipated by the Ice

The total heat dissipated by the ice is the sum Q1 + Q2. This total heat dissipated by the ice is equal to the heat lost by the water in the glass.

Calculate the Final Temperature of the Water

The heat lost by water in the glass can be calculated as Q = mcΔT. Now we can equate the heat lost by the water to the heat absorbed by the ice to find the final temperature Tf. That is, m_water*c_water*(T_initial-Tf) = Q1 + Q2. The specific heat capacity of water c_water = 4.18 J/g°C.

Repeat the Process for Part B

Repeat steps 1 to 4 to calculate the final temperature of the water when only one ice cube is used. In this case, the mass of the ice will be halved, which will influence Q1, Q2, and total heat dissipated by the ice.

Key concepts

Heat Transfer

Heat transfer, a cornerstone of thermodynamics, refers to the movement of energy from one place or material to another as a result of a temperature difference. It occurs in various forms such as conduction, convection, and radiation. When you place ice cubes into a glass of water, heat transfer begins immediately, with energy moving from the warmer water to the colder ice. This energy is not lost but is transferred, causing the ice to eventually melt and the water temperature to decrease until a state of thermal equilibrium is achieved, where both substances settle at the same temperature.

Conduction in Ice-Water System

Consider our melting ice scenario; here, conduction is the primary mode of heat transfer. Direct contact between the ice and water allows energy to flow until temperatures balance out. It's this principle that allows for the calculation of final drink temperatures in the given exercise.

Specific Heat Capacity

Specific heat capacity is a measure of how much heat energy is required to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). It's an intrinsic property that varies from one material to another. For instance, water has a relatively high specific heat capacity, meaning it can absorb or release a considerable amount of energy before changing temperature significantly. This is why our oceans can absorb much heat without a significant increase in temperature.

Importance in Calculations

In the context of our problem, specific heat capacity is crucial. It tells us how much energy the ice must absorb to reach 0°C from -15°C and also helps us understand the energy exchange as the water cools down. This property is key to calculating the final temperature of the water once the ice has melted.

Latent Heat of Fusion

The latent heat of fusion is the amount of energy required to change a substance from the solid phase to the liquid phase at its melting point without changing its temperature. This property is distinct from specific heat capacity because it describes a phase change rather than a temperature change. For water, this phase change happens at 0°C, and the latent heat of fusion for ice is significant, which means it absorbs a lot of energy to melt, energy that doesn't contribute to a temperature rise.

Phase Change in the Exercise

In the exercise, the latent heat of fusion is vital to quantify the energy needed to melt the ice cubes into water. This phase change must be factored into the total heat calculation because it represents a notable share of the energy exchange without altering temperature.

Temperature Change Calculations

Temperature change calculations involve determining how much a substance's temperature will alter when it gains or loses a certain amount of heat energy. In the exercise, we apply the concept to both ice heating from -15°C to 0°C and the subsequent cooling of water as the ice melts. The calculations use the formula Q = mcΔT for sensible heat, where 'Q' is the total heat energy, 'm' is the mass, 'c' is the specific heat capacity, and 'ΔT' is the change in temperature.

Calculating the Final Temperature

The final temperature is found by balancing the heat absorbed by the ice and the heat lost by the water. Using the specific heat capacities and masses, we can set up equations that, when solved, provide the final equilibrium temperature. This calculation is essential for both parts of the problem, whether melting two ice cubes or just one.