Problem 53 - University Physics with Modern Physics

Question

An insulated beaker with negligible mass contains 0.250 kg of water at 75.0\(^\circ\)C. How many kilograms of ice at \(-\)20.0\(^\circ\)C must be dropped into the water to make the final temperature of the system 40.0\(^\circ\)C?

Step-by-Step Solution

Answer

0.095 kg of ice is needed.

Step 1: Understanding the Problem

We need to find the mass of ice at -20.0°C required to cool down 0.250 kg of water from 75.0°C to 40.0°C, with the final system reaching equilibrium at 40.0°C. Energy exchange involves the heat lost by the water and the heat gained by the ice to reach equilibrium.

Step 2: Writing the Heat Exchange Equation

The heat lost by the water can be calculated using the formula: \\[ q_{water} = m_{water} \cdot c_{water} \cdot (T_{initial, water} - T_{final}) \] \For the ice, it has to first be warmed up to 0°C, melt, and then heat to 40°C: \\[ q_{ice} = m_{ice} \cdot c_{ice} \cdot (0 - T_{initial, ice}) + m_{ice} \cdot L_f + m_{ice} \cdot c_{water} \cdot (T_{final} - 0) \] \Here, \(c_{water} = 4.186 \, \text{J/g°C}\), \(c_{ice} = 2.09 \, \text{J/g°C}\), and \(L_f = 334 \, \text{J/g}\) are the specific heat capacities and latent heat of fusion, respectively.

Step 3: Calculating Heat Lost by Water

Using the equation for heat lost by water, we have: \\[ q_{water} = 0.250 \, \text{kg} \times 4.186 \, \text{kJ/kg°C} \times (75.0°C - 40.0°C) \] \Calculate this to find \(q_{water}\).

Step 4: Calculating Heat Gained by Ice

We calculate the three components of heat gained by ice: \1. Heating ice to 0°C: \\[ q_{1} = m_{ice} \cdot 2.09 \, \text{kJ/kg°C} \cdot (0 - (-20.0)) \] \2. Melting ice: \\[ q_{2} = m_{ice} \cdot 334 \, \text{kJ/kg} \] \3. Heating water from 0°C to 40°C: \\[ q_{3} = m_{ice} \cdot 4.186 \, \text{kJ/kg°C} \cdot (40.0 - 0) \]

Step 5: Setting Up the Equation

Set the heat lost by the water equal to the heat gained by the ice: \\[ q_{water} = q_{1} + q_{2} + q_{3} \] \Substitute the values calculated in previous steps into this equation.

Step 6: Solving the Equation for Mass of Ice

Substitute \(q_{water}\), \(q_{1}\), \(q_{2}\), and \(q_{3}\) into the equation and solve for \(m_{ice}\). This requires rearranging the equation to isolate \(m_{ice}\) on one side.

Key Concepts

Heat Transfer

In the context of reaching thermal equilibrium, heat transfer plays a fundamental role. Consider a system where water and ice interact within an insulated beaker. The principle governing this interaction is the transfer of heat. When objects at different temperatures come into contact, heat will naturally transfer from the warmer object to the cooler one until thermal equilibrium is reached.

In this specific exercise, the heat lost by the warm water is transferred to the colder ice. This process allows the ice to warm up from its initial sub-zero temperature, melt, and eventually reach the equilibrium temperature of 40.0°C. Understanding the nature of heat flow helps in estimating how much energy, or heat, is needed to change the state of the substances involved, and thus, calculating the precise mass of ice required.

Specific Heat Capacity

Specific heat capacity is a measure of how much energy is needed to change the temperature of a given mass of a substance by one degree Celsius. Different materials have different specific heat capacities, indicating some require more heat to change temperature than others.

In the beaker problem, we deal with ice and water. Water, with a specific heat capacity of 4.186 J/g°C, needs more energy to change its temperature compared to ice, which has a specific heat capacity of 2.09 J/g°C. These values reflect how much heat each substance can absorb or release when their temperatures change.

• For water: A higher specific heat capacity means it can absorb more heat without a significant change in temperature.
• For ice: A lower specific heat capacity means less heat is needed to change its temperature by each degree.

The distinction in specific heat capacities is vital for the calculations that determine how the temperature of the water and ice changes until they reach the same 40.0°C.

Latent Heat of Fusion

The latent heat of fusion is the energy needed for a substance to change state from solid to liquid without changing temperature. This concept is significant when considering the energy required to melt ice.

In melting, each gram of ice requires an exact amount of heat energy known as the latent heat of fusion. For water, this amount is 334 J/g, which is the energy needed to transform ice at its melting point into water without any temperature change.

During the exercise, once the ice warms up to 0°C, additional energy is needed to transform it from solid to liquid, signifying why the latent heat of fusion is critical. This ensures that the melting process is measured accurately, as this stage requires a substantial amount of the total heat gained by the ice before it can increase in temperature as liquid water. By accurately calculating the energy required for melting, students can determine how much ice is necessary for the whole system to settle at the desired equilibrium temperature.