Problem 12 - Master Resource Book in JEE Main Physics

Question

Which one of the following would raise the temperature of \(20 \mathrm{~g}\) of water at \(30^{\circ} \mathrm{C}\) most when mixed with it? (a) \(20 \mathrm{~g}\) of water at \(40^{\circ} \mathrm{C}\) (b) \(40 \mathrm{~g}\) of water at \(35^{\circ} \mathrm{C}\) (c) \(10 \mathrm{~g}\) of water at \(50^{\circ} \mathrm{C}\) (d) \(4 \mathrm{~g}\) of water at \(80^{\circ} \mathrm{C}\)

Step-by-Step Solution

Answer

Option (d) raises the temperature most, to 63.33°C.

Step 1: Identify the Mixing Formula

To solve this problem, we need to calculate the final temperature when different hot water samples are mixed with the initial water at a lower temperature. We will use the formula for heat exchange:\[m_1c(T_f - T_{i1}) + m_2c(T_f - T_{i2}) = 0\]where \(m_1\) and \(m_2\) are the masses, \(T_{i1}\) and \(T_{i2}\) the initial temperatures, and \(T_f\) is the final temperature. Here, \(c\) is the specific heat capacity of water, which we can ignore as it cancels out in calculations.

Step 2: Calculate Final Temperature for Option (a)

For option (a):- Masses: \(m_1 = 20\, \text{g}\), \(m_2 = 20\, \text{g}\)- Initial temperatures: \(T_{i1} = 30\, ^\circ\text{C}\), \(T_{i2} = 40\, ^\circ\text{C}\)Using the heat exchange formula:\[20(T_f - 30) + 20(T_f - 40) = 0\]Simplifying gives:\[40T_f - 1400 = 0\]\[T_f = 35\, ^\circ\text{C}\]

Step 3: Calculate Final Temperature for Option (b)

For option (b):- Masses: \(m_1 = 20\, \text{g}\), \(m_2 = 40\, \text{g}\)- Initial temperatures: \(T_{i1} = 30\, ^\circ\text{C}\), \(T_{i2} = 35\, ^\circ\text{C}\)Using the heat exchange formula:\[20(T_f - 30) + 40(T_f - 35) = 0\]Simplifying gives:\[60T_f - 2050 = 0\]\[T_f = 34.17\, ^\circ\text{C}\]

Step 4: Calculate Final Temperature for Option (c)

For option (c):- Masses: \(m_1 = 20\, \text{g}\), \(m_2 = 10\, \text{g}\)- Initial temperatures: \(T_{i1} = 30\, ^\circ\text{C}\), \(T_{i2} = 50\, ^\circ\text{C}\)Using the heat exchange formula:\[20(T_f - 30) + 10(T_f - 50) = 0\]Simplifying gives:\[30T_f - 1300 = 0\]\[T_f = 43.33\, ^\circ\text{C}\]

Step 5: Calculate Final Temperature for Option (d)

For option (d):- Masses: \(m_1 = 20\, \text{g}\), \(m_2 = 4\, \text{g}\)- Initial temperatures: \(T_{i1} = 30\, ^\circ\text{C}\), \(T_{i2} = 80\, ^\circ\text{C}\)Using the heat exchange formula:\[20(T_f - 30) + 4(T_f - 80) = 0\]Simplifying gives:\[24T_f - 1520 = 0\]\[T_f = 63.33\, ^\circ\text{C}\]

Step 6: Determine the Highest Final Temperature

Comparing all the calculated final temperatures:- Option (a): \(35\, ^\circ\text{C}\)- Option (b): \(34.17\, ^\circ\text{C}\)- Option (c): \(43.33\, ^\circ\text{C}\)- Option (d): \(63.33\, ^\circ\text{C}\)Option (d) gives the highest final temperature, \(63.33\, ^\circ\text{C}\).

Key Concepts

Specific Heat Capacity

When exploring the principles of heat exchange, one fundamental concept is specific heat capacity. This refers to the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius. For water, this value is quite high, at approximately 4.18 joules per gram per degree Celsius. This means that water can absorb or release a lot of heat without changing its temperature significantly. In many calculations or theoretical analyses, like the one in our exercise, we can ignore the specific heat capacity if it's consistent across all substances involved. In our water mixing scenarios, since every sample involves only water, the specific heat capacity cancels out when comparing the heat exchange balance.

Final Temperature Calculation

The formula used to determine the final temperature of a mixture is a balance of heat lost and gained. When two bodies at different temperatures are mixed, heat will transfer until thermal equilibrium is reached. The formula is expressed as follows:\[m_1c(T_f - T_{i1}) + m_2c(T_f - T_{i2}) = 0\]where:

• \(m_1\) and \(m_2\) are the masses of the two substances.
• \(T_{i1}\) and \(T_{i2}\) are their initial temperatures.
• \(T_f\) is the final equilibrium temperature.
• \(c\) is the specific heat capacity, which we often ignore if it's the same for both substances.

This equation assumes no heat loss to the environment, meaning all heat exchange is internal to the system. Simplifying this balance gives the temperature at which both bodies reach equilibrium. Each part of the equation represents the heat change of each component, and solving the equation allows us to find \(T_f\).

Mixing Water at Different Temperatures

When mixing bodies of water with different temperatures, the concept of heat exchange comes into play. The farther apart the initial temperatures, the greater the potential for heat exchange. This is because the hotter water will lose heat, which is absorbed by the cooler water until a balanced temperature is reached. Several factors affect the final temperature:

• Initial temperatures: The greater the difference, the higher the heat flow potential.
• Masses of water: A larger mass may have a significant impact as it can store more heat energy.

In our specific exercise, we looked at different scenarios of mixing water at various initial temperatures and calculated the final temperatures using the mass and temperature inputs for each option. The choice with the smallest original mass, but highest initial temperature, resulted in the highest final temperature due to the substantial heat it could impart.