Problem 48 - College 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 in the water so that the final temperature of the system will be \(30.0^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

The required mass of ice is approximately 0.094 kg.

Step 1: Identify the Known Variables

The mass of the water, \( m_w = 0.250 \) kg, its initial temperature, \( T_{w_i} = 75.0^{\circ} \mathrm{C} \), and the specific heat capacity of water, \( c_w = 4186 \, \mathrm{J/kg^{\circ}C} \). The ice has an initial temperature of \( T_{i_i} = -20.0^{\circ} \mathrm{C} \) and needs to reach a final temperature of \( T_f = 30.0^{\circ} \mathrm{C} \) just like the water.

Step 2: Calculate Heat Released by Water

Calculate the heat released by the water as it cools from \( 75.0^{\circ} \mathrm{C} \) to \( 30.0^{\circ} \mathrm{C} \). Use the formula:\[ Q_{w} = m_w c_w (T_{f} - T_{w_i}) = 0.250 \times 4186 \times (30 - 75) \].Calculate \( Q_w = -46992 \, \mathrm{J} \). The negative sign indicates heat release.

Step 3: Calculate Heat Needed to Warm Ice to 0°C

Calculate the heat required to raise the temperature of ice from \( -20.0^{\circ} \mathrm{C} \) to \( 0^{\circ} \mathrm{C} \). Use the specific heat capacity of ice, \( c_i = 2090 \, \mathrm{J/kg^{\circ}C} \):\[ Q_{1} = m c_i (0 - (-20)) = m \times 2090 \times 20 \].This simplifies to \( Q_{1} = 41800m \).

Step 4: Calculate Heat Needed to Melt Ice

Calculate the heat required to melt the ice at \( 0^{\circ} \mathrm{C} \) into water. Use the latent heat of fusion for ice, \( L_f = 334000 \, \mathrm{J/kg} \):\[ Q_2 = m L_f = m \times 334000 \].

Step 5: Calculate Heat Needed to Warm Melted Ice to Final Temperature

Calculate the heat required to warm the resulting water (from melted ice) from \( 0^{\circ} \mathrm{C} \) to \( 30.0^{\circ} \mathrm{C} \):\[ Q_3 = m c_w (30 - 0) = m \times 4186 \times 30 \].This simplifies to \( Q_3 = 125580m \).

Step 6: Set Up the Energy Balance Equation

The total heat absorbed by the ice must equal the heat released by the water. Set up the equation:\[ Q_1 + Q_2 + Q_3 = -Q_w \] \[(41800m) + (334000m) + (125580m) = 46992 \].

Step 7: Solve for the Mass of Ice

Combine and solve for \( m \):\[ 501380m = 46992 \] \[ m = \frac{46992}{501380} \approx 0.0937 \mathrm{~kg} \].

Key Concepts

Specific Heat Capacity

When discussing heat transfer, specific heat capacity plays a crucial role. It defines how much heat energy is needed to raise the temperature of a unit mass of a substance by one degree Celsius. This property varies between different materials. For example, in the original exercise, water's specific heat capacity is given as \( c_w = 4186 \, \mathrm{J/kg^{\circ}C} \). This means it takes 4186 Joules of energy to increase the temperature of 1 kg of water by 1°C.
This concept is essential when calculating how much heat energy a substance (like water) releases or absorbs as it changes temperature without changing its state.

Latent Heat of Fusion

The latent heat of fusion refers to the energy required to change a substance from a solid to a liquid at its melting point, without changing its temperature. For ice, this is a significant factor in the calculation of energy exchange. In the exercise, the latent heat of fusion for ice is specified as \( L_f = 334000 \, \mathrm{J/kg} \).
This value indicates that 334,000 Joules are needed to melt 1 kg of ice at 0°C into water at the same temperature.
Even though the temperature remains constant during this phase change, a substantial amount of energy is required to overcome the forces holding the particles in a solid phase.

Thermal Equilibrium

Thermal equilibrium is a crucial concept in the analysis of heat transfer exercises. It occurs when two or more objects reach a common temperature due to energy exchange, with no net heat flow between them.
In our example, the system reaches thermal equilibrium when the ice completely melts and the water and resulting water from ice have settled at a final temperature of 30°C.
This balance is achieved because the heat lost by the warmer water matches the heat gained by the ice as it melts and warms up.

Energy Conservation

Energy conservation is a fundamental principle that applies to every physical interaction, including heat transfer. It states that energy cannot be created or destroyed, only transformed from one form to another.
In the context of our ice melting problem, energy conservation implies that the total energy released by the cooling water must equal the total energy absorbed by the ice as it warms, melts, and continues to warm as water.

• The heat lost by the water as it cools is calculated with its specific heat capacity.
• The energy needed to warm, melt, and further warm the ice is calculated with the specific heat capacity of ice and the latent heat of fusion.

Solving the energy balance equation ensures we correctly apply energy conservation to find the mass of ice required.