Problem 54 - Arihant AIEEE Physics

Question

\(5 \mathrm{~g}\) of water at \(30^{\circ} \mathrm{C}\) and \(5 \mathrm{~g}\) of ice at \(-20^{\circ} \mathrm{C}\) are mixed together in a calorimeter. The water equivalent of calorimeter is negligible and specific heat and latent heat of ice are \(0.5 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}\) and \(80 \mathrm{cal} / \mathrm{g}\) respectively. The final temperature of the mixure is: (a) \(0^{\circ} \mathrm{C}\) (b) \(-8^{\circ} \mathrm{C}\) (c) \(-4^{\circ} \mathrm{C}\) (d) \(2^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

(a) The final temperature is \(0^{\circ} \mathrm{C}\).

Step 1: Calculate the heat required to warm ice to 0°C

To find the heat required to raise the temperature of ice from \(-20^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\), we use the formula for heat transfer \(q = mc\Delta T\), where \(m\) is the mass, \(c\) is the specific heat, and \(\Delta T\) is the change in temperature. Here, \(m = 5\, \mathrm{g}\), \(c = 0.5\, \mathrm{cal/g}^{\circ}\,\mathrm{C}\), and \(\Delta T = 20^{\circ}\,\mathrm{C}\). Thus,\[q = 5 \, \mathrm{g} \times 0.5 \, \mathrm{cal/g}^{\circ} \mathrm{C} \times 20^{\circ} \mathrm{C} = 50 \, \mathrm{cal}\].

Step 2: Calculate the heat required to melt ice at 0°C

The latent heat of fusion is needed to convert the ice at \(0^{\circ} \mathrm{C}\) to water. The formula for this is \(q = mL\), where \(L\) is the latent heat of fusion of ice. Here, \(m = 5\, \mathrm{g}\) and \(L = 80\, \mathrm{cal/g}\), so\[q = 5 \, \mathrm{g} \times 80 \, \mathrm{cal/g} = 400 \, \mathrm{cal}\].

Step 3: Calculate the heat released by water cooling to 0°C

The heat released by the water as it cools down from \(30^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\) is calculated with the formula \(q = mc \Delta T\), where \(m = 5\, \mathrm{g}\), \(c = 1 \, \mathrm{cal/g}^{\circ} \mathrm{C}\), and \(\Delta T = 30^{\circ} \mathrm{C}\). Thus,\[q = 5 \, \mathrm{g} \times 1 \, \mathrm{cal/g}^{\circ} \mathrm{C} \times 30^{\circ} \mathrm{C} = 150 \, \mathrm{cal}\].

Step 4: Determine if ice fully melts with available heat

The heat released by the water (150 cal) is less than the total heat required to both warm the ice to \(0^{\circ} \mathrm{C}\) \[(50 \, \mathrm{cal})\] and melt it to water \[(400 \, \mathrm{cal})\]. Since only 150 cal is available, some ice remains unmelted.

Step 5: Evaluate the equilibrium temperature

Since not enough heat is available to melt all ice, the equilibrium temperature will be at a state where water and ice coexist. The final temperature is at \(0^{\circ} \mathrm{C}\) because partial melting and temperature rise of ice until this point consume all the available heat.

Key Concepts

Heat Transfer

Heat transfer is the process of energy moving from a warmer object to a cooler one. In calorimetry, this is a central idea as it helps determine how substances reach equilibrium. In our example, the heat from the warmer water needs to be transferred to the cooler ice. This transfer process is guided by the formula:

• For the water losing heat: \( q = mc \Delta T \) where \( q \) is heat, \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the change in temperature.
• For the ice gaining heat: it must first warm up to its melting point, using \( q = mc \Delta T \), and then melt, using the latent heat formula \( q = mL \).

The overall balance of heat transferred will decide if all ice can melt or if some will stay intact. This interplay is an elegant dance dictated by the available heat in the system.

Latent Heat

Latent heat deals with phase changes of substances without any change in temperature. This is crucial when ice melts to water. For our problem, we consider latent heat of fusion, which is the heat required to convert ice into water at the same temperature, for example, at 0°C. In practical terms:

• Latent heat of fusion for ice is \(80\, \text{cal/g}\).
• The ice requires \(5\, \text{g} \times 80\, \text{cal/g} = 400\, \text{cal}\) to completely melt.

This means even if you have heated the ice, it won't raise in temperature until all of it has turned into water due to the consumption of energy in undergoing phase change. This highlights how critical latent heat is in calculations involving changes of state, alongside other heating processes.

Specific Heat Capacity

Specific heat capacity is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. It varies between substances, influencing how they react to adding or removing heat. In this scenario, two key specific heat values apply:

• For ice, \(c = 0.5\, \text{cal/g°C}\).
• For water, \(c = 1\, \text{cal/g°C}\).

Notice the value of water is doubled. This means per gram, more energy is required to change water's temperature compared to ice. Specifically in our exercise, the initial warming of the ice (from \(-20^{\circ} \text{C}\) to \(0^{\circ} \text{C}\)) involves calculating how much energy is needed. This concept is sat at the heart of understanding why the heat dynamics is pivotal in determining final states in thermal equilibrium.

Thermal Equilibrium

Thermal equilibrium is achieved when no more heat flows between substances in contact. Essentially, they reach the same temperature. In our exercise, water and ice try reaching it using the heat transfer process described above. Initially, the water and ice are far from equilibrium due to their temperature differences:

• The water starts at \(30^{\circ} \text{C}\).
• The ice starts much colder at \(-20^{\circ} \text{C}\).

Through heat transfer, the water cools, giving off heat, which then allows ice to warm and begin to melt. If there isn't enough heat for all the ice to melt, they settle at an equilibrium point of \(0^{\circ} \text{C}\). This is the case here where ice remains, indicating a balance between the two phases, water at freezing point and unmelted ice, both coexisting harmoniously. Knowing this principle of reaching equilibrium helps to determine the final state of combined substances in thermodynamic processes.