Problem 53 - Master Resource Book in JEE Main Physics

Question

\(2 \mathrm{~kg}\) of ice at \(-20^{\circ} \mathrm{C}\) is mixed with \(5 \mathrm{~kg}\) of water at \(20^{\circ} \mathrm{C}\) in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water and ice are \(1 \mathrm{kcal} / \mathrm{kg}\) per \(^{\circ} \mathrm{C}\) and \(0.5 \mathrm{kcal} / \mathrm{kg} /{ }^{\circ} \mathrm{C}\) while the latent heat of fusion of ice is \(80 \mathrm{kcal} / \mathrm{kg}\). (a) \(7 \mathrm{~kg}_{1}\) (b) \(6 \mathrm{~kg}\) (c) \(4 \mathrm{~kg}\) (d) \(2 \mathrm{~kg}\)

Step-by-Step Solution

Answer

The final mass of water is 6 kg.

Step 1: Calculate Heat Required to Raise Ice to 0°C

First, calculate how much heat is needed to bring the ice at \(-20^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\). The formula to use is \(Q = mc\Delta T\), where \(m\) is the mass, \(c\) is the specific heat, and \(\Delta T\) is the change in temperature.For the ice, \(m = 2 \mathrm{~kg}\),\(c = 0.5 \mathrm{kcal} / \mathrm{kg}/^{\circ} \mathrm{C}\), and \(\Delta T = 0 - (-20) = 20^{\circ}\mathrm{C}\).So, the heat required is: \[ Q = 2 \times 0.5 \times 20 = 20 \mathrm{kcal} \]

Step 2: Calculate Heat Required to Melt the Ice

Next, calculate the heat required to melt the ice at \(0^{\circ} C\). This uses the latent heat of fusion: \(Q = mL\), where \(m = 2 \mathrm{~kg}\) and \(L = 80 \mathrm{kcal}/\mathrm{kg}\).The heat required to melt the ice is: \[ Q = 2 \times 80 = 160 \mathrm{kcal} \]

Step 3: Total Heat Needed for Ice to Become Water

The total heat required to bring \(2 \mathrm{~kg}\) of ice at \(-20^{\circ} \mathrm{C}\) to waterat \(0^{\circ} \mathrm{C}\) is the sum of steps 1 and 2:\[ Q_{\text{total}} = 20 + 160 = 180 \mathrm{kcal} \]

Step 4: Calculate Heat Released by Water

Calculate the heat released by \(5 \mathrm{~kg}\) of water cooling from \(20^{\circ} \mathrm{C}\) to \(0^{\circ} \mathrm{C}\).Use the formula \(Q = mc\Delta T\), where \(m = 5 \mathrm{~kg}\), \(c = 1 \mathrm{kcal} / \mathrm{kg}/^{\circ} \mathrm{C}\),and \(\Delta T = 0 - 20 = -20^{\circ}\mathrm{C}\).The heat released is: \[Q = 5 \times 1 \times 20 = 100 \mathrm{kcal} \]

Step 5: Determine if Ice Completely Melts

Compare the heat available from the water (100 kcal) with the heat required to convert ice to water (180 kcal). Since 100 kcal is less than 180 kcal, not all the ice will melt and some will remain as ice.

Step 6: Calculate Final Mass of Water

Since not all the ice melts, calculate how much ice melts using the available 100 kcal heat.Calculate the mass of ice melted using the heat available:For the ice at \(0^{\circ} C\), the energy required to melt it is given by:\[ mL = 100 \] with \(L = 80 \mathrm{kcal}/\mathrm{kg}\),\[ m = \frac{100}{80} = 1.25 \mathrm{~kg} \]So, 1.25 kg of ice melts, converting to water.

Step 7: Final Mass of Remaining Water

Add the water formed from the melted ice to the original water mass:The water from melted ice is 1.25 kg, and the original water mass is 5 kg.Hence, total water in the container is: \[ 5 + 1.25 = 6.25 \mathrm{~kg} \]

Step 8: Round to Nearest Option

Round the calculated mass (6.25 kg) to match the options given. The closest option is 6 kg, so select that as the final answer.

Key Concepts

Specific Heat Capacity

Specific heat capacity is a measure of how much heat energy is needed to raise the temperature of a substance by 1°C. It tells us how resistant a material is to changes in temperature.
For example, ice at -20°C has a specific heat capacity of 0.5 kcal/kg/°C, meaning it requires 0.5 kcal to raise 1 kg of ice by 1°C.
Meanwhile, water's specific heat capacity is 1 kcal/kg/°C, indicating it can absorb more heat energy compared to ice for each degree of temperature increase.
In simple terms, it shows why ice needs less heat to warm compared to water, due to its lower specific heat capacity.

Latent Heat of Fusion

Latent heat of fusion is the amount of energy needed to change a solid into a liquid at a constant temperature. It's a crucial concept when discussing phase changes, like melting ice into water.
For ice, the latent heat of fusion is 80 kcal/kg, which is the energy needed to melt 1 kg of ice at 0°C.
This energy is used to break the bonds holding the ice together as a solid, rather than raising its temperature. That's why when melting ice, you don't see a temperature change until all ice has turned to water.

Phase Change

A phase change occurs when a substance transitions from one state of matter to another, such as from solid to liquid or liquid to gas.
These changes happen without a change in temperature, as seen when ice melts, where heat is absorbed but the temperature remains constant until all ice is liquid.
During this process, heat energy is used to break molecular bonds rather than increase kinetic energy (temperature). This explains why phase changes are energy-intensive without corresponding temperature change.

Thermodynamics

Thermodynamics is the branch of physics that deals with heat and temperature and their relation to energy and work. It encompasses concepts like specific heat capacity and latent heat of fusion.
In this scenario, thermodynamics helps us understand how energy is transferred between water and ice, predicting the final state of the system.
The principles of thermodynamics guide calculations, such as the heat balance between cooling water and warming ice in insulated environments.

Temperature Change

Temperature change refers to the difference in temperature observed in a substance due to the gain or loss of heat energy.
In the given exercise, the ice increases from -20°C to 0°C, while the water decreases from 20°C to 0°C.
The formula for calculating the energy change due to temperature change is: \[ Q = mc\Delta T \] where \( Q \) is heat energy, \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the temperature change. This helps quantify the energy transfers required for a given temperature adjustment.