Problem 45 - A Complete Resource Book in Chemistry for JEE Main

Question

The molar heat capacity of water at constant pressure, C , is \(75 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}\). When \(1.0 \mathrm{~kJ}\) of heat is supplied to \(100 \mathrm{~g}\) of water which is free to expand, the increase in temperature of water is (a) \(4.8 \mathrm{~K}\) (b) \(6.6 \mathrm{~K}\) (c) \(1.2 \mathrm{~K}\) (d) \(2.4 \mathrm{~K}\)

Step-by-Step Solution

Answer

The increase in temperature is 2.4 K, option (d).

Step 1: Determine moles of water

First, we need to calculate the number of moles of water present in 100 g of water. The molar mass of water is approximately 18 g/mol. The number of moles, \( n \), is calculated using the formula:\[n = \frac{\text{mass}}{\text{molar mass}} = \frac{100 \text{ g}}{18 \text{ g/mol}} \approx 5.56 \text{ mol}\]

Step 2: Set up the formula for heat transfer

To find the increase in temperature, we use the formula for heat transfer at constant pressure, given by:\[\Delta Q = n \cdot C \cdot \Delta T\]Where \( \Delta Q \) is the heat supplied, \( n \) is the number of moles, \( C \) is the molar heat capacity, and \( \Delta T \) is the change in temperature we need to find.

Step 3: Input values and calculate temperature change

We have \( \Delta Q = 1000 \text{ J} \), \( n = 5.56 \text{ mol} \), and \( C = 75 \text{ J/K mol} \). Plug these values into the equation:\[1000 = 5.56 \cdot 75 \cdot \Delta T\]Solve for \( \Delta T \):\[\Delta T = \frac{1000}{5.56 \cdot 75} \approx 2.40 \text{ K}\]

Step 4: Select the correct answer

Based on our calculation, the increase in temperature of the water is approximately 2.40 K. Therefore, the correct answer is (d) 2.4 K.

Key Concepts

Heat Transfer

Heat transfer occurs when thermal energy is exchanged between objects or systems. This can happen through different processes such as conduction, convection, or radiation. For this exercise, we focus on the heat transfer at constant pressure, which means the water is allowed to expand freely as it absorbs heat.
In this context, the amount of heat transferred to the water, noted as \( \Delta Q \), is given as 1000 Joules (or 1 kJ). The relation between heat transfer and temperature change is crucial in determining how the temperature of a substance changes when it receives a certain amount of heat.

• Remember that heat capacity is a measure of how much heat energy is needed to raise the temperature of a given amount of substance by one degree Celsius or one Kelvin.
• Molar heat capacity is specifically about heat needed per mole of substance, which in this case is for water.
• We use the formula \( \Delta Q = n \cdot C \cdot \Delta T \) to relate these elements, where \( n \) is moles, \( C \) is molar heat capacity, and \( \Delta T \) is temperature change.

Understanding the relationship among these variables is key to predicting and calculating how substances behave with heat transfer.

Mole Calculation

Mole calculation is fundamental when dealing with chemical substances because it provides a bridge between the macroscopic measurements we make and the microscopic quantities we calculate for atoms and molecules.
Moles tell us how many molecules or atoms are present. In this problem, we need to know how many moles of water are in 100 grams so we can understand how it will react to the supplied heat.
To find the moles, we use the formula:

• \( n = \frac{\text{mass}}{\text{molar mass}} \)
• Here, the mass is 100 grams and the molar mass of water is around 18 g/mol.
• Calculate as follows: \( n = \frac{100}{18} \approx 5.56 \) moles.

This value of 5.56 moles is used in our heat transfer equation to determine how much the temperature of the water will increase under the given conditions.

Temperature Change

Temperature change is the final piece in understanding how heat affects a given amount of a substance. Here, it represents how much warmer the water gets when a certain amount of heat is absorbed.
To find this change, we solve for \( \Delta T \) in the heat transfer equation \( \Delta Q = n \cdot C \cdot \Delta T \).
By using the values: heat supplied \( \Delta Q = 1000 \) Joules, number of moles \( n = 5.56 \), and the molar heat capacity \( C = 75 \) J/K mol, we can rearrange and solve:

• \( \Delta T = \frac{1000}{5.56 \cdot 75} \approx 2.40 \) Kelvin.

This value of \( \Delta T \) means the water's temperature increases by 2.4 Kelvin when exposed to the given conditions and confirms that option (d) is correct.