Problem 1068 - JEE Physics

Question

A vessel contains \(110 \mathrm{~g}\) of water the heat capacity of the vessel is equal to \(10 \mathrm{~g}\) of water. The initial temperature of water in vessel is \(10^{\circ} \mathrm{C}\) If \(220 \mathrm{~g}\) of hot water at \(70^{\circ} \mathrm{C}\) is poured in the vessel the Final temperature neglecting radiation loss will be (A) \(70^{\circ} \mathrm{C}\) (B) \(80^{\circ} \mathrm{C}\) (C) \(60^{\circ} \mathrm{C}\) (D) \(50^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

The final temperature of the water when 220 grams of hot water at 70°C is poured into a vessel containing 110 grams of water at 10°C, neglecting radiation loss, is approximately \(49^{\circ} \mathrm{C}\), which is closest to option (D) \(50^{\circ} \mathrm{C}\).

Step 1: Set up the equation for heat transfer

As we stated earlier, heat gained by the cooler water is equal to the heat lost by the hotter water. We can set up an equation for this. \(Q_{gained} = Q_{lost}\)

Step 2: Define the variables

Now we will define the variables for the masses and temperatures of water: \(m_1 = 110\) g (mass of the cooler water) \(m_2 = 220\) g (mass of the hotter water) \(T_1 = 10\) °C (initial temperature of the cooler water) \(T_2 = 70\) °C (initial temperature of the hotter water) \(C = 1\) cal/g·°C (specific heat capacity of water) \(V = 10\) g (the heat capacity of the vessel is equal to this) Now, the equation for heat transfer will involve the specific heat capacity of water.

Step 3: Express heat gained by the cooler water

Heat gained by the cooler water (including the vessel) can be expressed as: \(Q_{gained} = (m_1 + V) * C * (T_f - T_1)\)

Step 4: Express heat lost by the hotter water

Heat lost by the hotter water can be expressed as: \(Q_{lost} = m_2 * C * (T_2 - T_f)\)

Step 5: Set up the equation and solve for final temperature

Now we can set up the equation with all gathered data. \((m_1 + V) * C * (T_f - T_1) = m_2 * C * (T_2 - T_f)\) As \(C\) is present on both sides of the equation, it can be cancelled out. The equation becomes: \((110 + 10) (T_f - 10) = 220 (70 - T_f)\) Solving for \(T_f\): \(120(T_f - 10) = 220(70 - T_f)\) Now, expand and simplify the equation: \(120T_f - 1200 = 15400 - 220T_f\) Combine like terms: \(340T_f = 16600\) Now, divide both sides by 340: \(T_f = 48.8235294\) Since we need the final temperature in whole numbers, we can round up the final value to the nearest whole number \(T_f \approx 49^{\circ} \mathrm{C}\) The final temperature is between \(50^{\circ} \mathrm{C}\) and \(60^{\circ} \mathrm{C}\), which means our answer is closest to option (D) \(50^{\circ} \mathrm{C}\).

Key Concepts

Specific Heat Capacity

Specific heat capacity is a crucial concept in understanding how different substances absorb and transfer heat. It represents the amount of heat per unit mass required to raise the temperature of a substance by one degree Celsius. For water, this value is fairly high at 1 calorie per gram per degree Celsius (cal/g·°C). This means that water can absorb a lot of heat before its temperature rises significantly, making it an excellent medium for heat exchange.

In the provided problem, specific heat capacity helps us determine how much heat the cooler water and vessel will absorb when mixed with the hotter water. By using the same specific heat capacity for calculations, simplifications in the mathematical process become possible, enabling us to easily compute the final temperature of the mixture.

Temperature Equilibrium

Temperature equilibrium is the state where two or more objects in contact reach a shared temperature. An important principle of thermal dynamics is that heat flows from the hotter object to the cooler one until equilibrium is achieved. This process is governed by the principle of conservation of energy. No heat is lost or gained overall; it simply moves between substances.

In our problem, when the hot water is added to the cooler water and vessel, the final temperature is reached when the heat lost by the hot water equals the heat gained by the cooler water and the vessel. By setting up an equation where these two quantities are equal, we can solve for the temperature equilibrium, showcasing the predictable balance of heat transfer dynamics. This concept is indispensable in solving real-world problems involving mixtures of different temperature substances.

Heat Capacity of Materials

The heat capacity of a material defines how much heat is needed to change its temperature. Unlike specific heat capacity, which is measured per unit mass, heat capacity is an extensive property that depends on the entire object. For instance, a vessel's heat capacity could vary drastically based on its size, shape, and material composition.

In this exercise, the vessel's heat capacity is given as equivalent to 10 grams of water. This means the vessel itself will absorb heat similarly to 10 grams of water. Such considerations are vital, especially when dealing with experiments or calculations involving not just the contents but the container itself. Understanding the heat capacities of different materials allows scientists and engineers to make precise predictions about how systems will respond to thermal energy changes.