To solve this problem, we need to apply the principle of conservation of energy, which states that the heat lost by the warmer substance (water) will equal the heat gained by the colder substance (ice). Let's break this down step by step for both scenarios: two ice cubes and one ice cube.
Understanding the Heat Transfer
When the ice cubes are added to the water, they will absorb heat from the water until thermal equilibrium is reached, meaning both the water and the ice will be at the same final temperature. We will use the specific heat capacities of water and ice to calculate the final temperature.
Key Values
- Specific heat capacity of water: 4.18 J/g°C
- Specific heat capacity of ice: 2.09 J/g°C
- Latent heat of fusion of ice: 334 J/g
Scenario (a): Two Ice Cubes
We have two ice cubes, each weighing 50 g, so the total mass of ice is 100 g. The water mass is 200 g, and the initial temperatures are 25ºC for the water and -15ºC for the ice. We need to calculate how much heat is required to bring the ice to 0ºC, then to melt it, and finally to find the final temperature of the mixture.
Step 1: Heating the Ice to 0ºC
The heat required to raise the temperature of the ice from -15ºC to 0ºC can be calculated using the formula:
Q1 = m * c * ΔT
Where:
- m = mass of ice = 100 g
- c = specific heat capacity of ice = 2.09 J/g°C
- ΔT = change in temperature = 0 - (-15) = 15°C
Calculating Q1:
Q1 = 100 g * 2.09 J/g°C * 15°C = 3135 J
Step 2: Melting the Ice
Next, we need to calculate the heat required to melt the ice at 0ºC:
Q2 = m * Lf
Where:
- m = mass of ice = 100 g
- Lf = latent heat of fusion = 334 J/g
Calculating Q2:
Q2 = 100 g * 334 J/g = 33400 J
Step 3: Total Heat Absorbed by Ice
The total heat absorbed by the ice is:
Q_total_ice = Q1 + Q2 = 3135 J + 33400 J = 36535 J
Step 4: Heat Lost by Water
The heat lost by the water as it cools down can be expressed as:
Q_water = m * c * ΔT
Where:
- m = mass of water = 200 g
- c = specific heat capacity of water = 4.18 J/g°C
- ΔT = change in temperature (final temperature - initial temperature)
Let’s denote the final temperature as T_f. The heat lost by the water is:
Q_water = 200 g * 4.18 J/g°C * (T_f - 25°C)
Step 5: Setting Up the Equation
Since the heat lost by the water equals the heat gained by the ice, we can set up the equation:
200 g * 4.18 J/g°C * (T_f - 25°C) = 36535 J
Step 6: Solving for T_f
Now we can solve for T_f:
836 J/°C * (T_f - 25) = 36535
T_f - 25 = 36535 / 836
T_f - 25 = 43.7
T_f = 68.7°C
Scenario (b): One Ice Cube
If we only use one ice cube (50 g), we will repeat the calculations but with half the mass of ice.
Step 1: Heating the Ice to 0ºC
For one ice cube:
Q1 = 50 g * 2.09 J/g°C * 15°C = 1567.5 J
Step 2: Melting the Ice
Q2 = 50 g * 334 J/g = 16700 J
Step 3: Total Heat Absorbed by Ice
Q_total_ice = Q1 + Q2 = 1567.5 J + 16700 J = 18267.5 J
Step 4: Heat Lost by Water
Using the same formula for heat lost by water:
Q_water = 200 g * 4.18 J/g°C * (T_f - 25°C)
Step 5: Setting Up the Equation
200 g * 4.18 J/g°C * (T_f - 25°C) = 18267.5 J
Step 6: Solving for T_f
836 J/°C * (T_f - 25) = 18267.5
T_f - 25 = 18267.5 / 836
T_f - 25 = 21.85
T_f = 46.85°C
Final Results
In summary:
- With two ice cubes, the final temperature of the drink is approximately 68.7°C.
- With one ice cube, the final temperature of the drink is approximately 46.85°C.
This demonstrates how the amount of ice significantly affects the final temperature of the mixture due to the heat absorption capacity of the ice. The more ice you have, the more heat it can absorb, leading to a lower final temperature.
