Problem 59 - University Physics with Modern Physics

Question

An insulated beaker with negligible mass contains 0.250 \(\mathrm{kg}\) of water at a temperature of \(75.0^{\circ} \mathrm{C}\) . How many kilograms of ice at a temperature of \(-20.0^{\circ} \mathrm{C}\) must be dropped into the water to make the final temperature of the system \(30.0^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

Approximately 0.0945 kg of ice is needed.

Step 1: Identify the energy change in the water

The water initially has a mass of 0.250 kg and a temperature of 75.0°C. It cools down to 30.0°C. We calculate the energy lost by the water using the specific heat capacity of water, which is 4,186 J/kg°C. \[Q_{water} = m_{water} \cdot c_{water} \cdot \Delta T\]\[Q_{water} = 0.250 \times 4186 \times (30.0 - 75.0)\]\[ Q_{water} = 0.250 \times 4186 \times (-45) = -47,092.5 \, \text{J} \]

Step 2: Calculate the energy required to warm the ice to 0°C

The ice, initially at -20.0°C, must first be raised to 0°C. We use the specific heat capacity of ice, which is 2,090 J/kg°C.\[Q_{1} = m_{ice} \cdot c_{ice} \cdot \Delta T\]\[Q_{1} = m_{ice} \times 2090 \times (0 - (-20))\]\[Q_{1} = m_{ice} \times 2090 \times 20 = 41800 \, m_{ice} \, \text{J} \]

Step 3: Calculate the energy required to melt the ice

After the ice reaches 0°C, it needs to be melted. The latent heat of fusion for ice is 334,000 J/kg.\[Q_{2} = m_{ice} \cdot L_f\]\[Q_{2} = m_{ice} \times 334,000\]

Step 4: Calculate the energy required to warm the melted ice to 30°C

Once the ice has melted, the resulting water needs to be warmed from 0°C to 30°C. We use the specific heat capacity of water (4,186 J/kg°C) for this calculation.\[Q_{3} = m_{water} \cdot c_{water} \cdot \Delta T\]\[Q_{3} = m_{ice} \times 4186 \times 30\]

Step 5: Set up the energy balance equation

The total heat gained by the ice should equal the heat lost by the water. Thus, \[Q_{1} + Q_{2} + Q_{3} = -Q_{water}\]Substituting the expressions for \(Q_{1}\), \(Q_{2}\), and \(Q_{3}\) from steps 2, 3, and 4:\[41800 \times m_{ice} + 334000 \times m_{ice} + 125580 \times m_{ice} = 47092.5\]

Step 6: Solve for the mass of the ice

Combine and simplify the equation from step 5:\[41800 \times m_{ice} + 334000 \times m_{ice} + 125580 \times m_{ice} = 47092.5\]Combine like terms:\[498380 \times m_{ice} = 47092.5\]Solve for \(m_{ice}\):\[m_{ice} = \frac{47092.5}{498380} \approx 0.0945 \, \text{kg} \]

Key Concepts

Specific Heat Capacity

The specific heat capacity is an essential concept in thermodynamics. It refers to the amount of heat energy required to change the temperature of a unit mass of a substance by one degree Celsius. This property is crucial for understanding how different materials react to changes in temperature. In this exercise, the specific heat capacity of water is important because it helps us calculate how much energy is lost when water cools down from 75°C to 30°C.

For water, this value is 4,186 J/kg°C. The formula to find the energy change due to temperature change is: \( Q = m \cdot c \cdot \Delta T \), where:

• \( Q \) is the heat energy in joules (J)
• \( m \) is the mass in kilograms (kg)
• \( c \) is the specific heat capacity (J/kg°C)
• \( \Delta T \) is the change in temperature in degrees Celsius (°C)

This formula allows us to calculate how much energy the 0.250 kg of water at 75°C releases as it cools to 30°C.

Latent Heat of Fusion

The latent heat of fusion is the energy required to change a solid into a liquid without changing its temperature. It's a critical concept for situations where a substance transitions between solid and liquid states, such as in this exercise where ice melts into water.

For ice, the latent heat of fusion is 334,000 J/kg. This means it takes a substantial amount of energy to transform ice at 0°C into water at the same temperature. When calculating the energy needed to melt ice, the formula used is: \( Q = m \cdot L_f \), where:

• \( Q \) is the heat energy in joules (J)
• \( m \) is the mass of the ice (kg)
• \( L_f \) is the latent heat of fusion for the substance (J/kg)

No temperature change occurs during this process, highlighting the importance of understanding latent heat when involving phase changes.

Energy Balance Equation

The energy balance equation is fundamental in thermodynamics, ensuring the conservation of energy within a system. It dictates that the total energy gained by one part of the system must be equal to the energy lost by another part.

In our example, the energy lost by the water is gained by the ice. This means the cooling of water and the warming, melting, and further heating of ice must exactly exchange the same amount of energy. The equation is expressed as:\[ Q_{1} + Q_{2} + Q_{3} = -Q_{water} \]where:

• \( Q_{1} \) is the energy to warm the ice to 0°C
• \( Q_{2} \) is the energy to melt the ice
• \( Q_{3} \) is the energy to warm the melted ice to 30°C
• \( -Q_{water} \) is the energy lost by the water as it cools

Understanding this balance lets us solve for unknowns, such as the mass of ice needed to achieve our desired final system temperature.