The molar heat capacity of water at constant pressure, \(C_{P}\) is \(75 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}\). When \(10 \mathrm{~kJ}\) of heat is supplied to \(1 \mathrm{~kg}\) water which is free to expand, the increase in temperature of water is (a) \(2.4 \mathrm{~K}\) (b) \(4.8 \mathrm{~K}\) (c) \(3.2 \mathrm{~K}\) (d) \(10 \mathrm{~K}\)

Short Answer

Expert verified
The increase in temperature of water is \(4.8 \mathrm{~K}\).

Step by step solution

01

Convert Heat Supplied to Joules

The first step is to convert the heat supplied to the water from kilojoules to joules. Since 1 kilojoule equals 1000 joules, we can do this by multiplying the energy in kilojoules by 1000.
02

Calculate the Moles of Water

Next, calculate the number of moles of water. We know that the molar mass of water is approximately 18.015 \(\mathrm{g/mol}\). To find the number of moles, divide the mass of the water by the molar mass.
03

Calculate Temperature Increase

Use the formula for heat transfer at constant pressure: \(q = nC_P\Delta T\), where \(q\) is the heat transferred, \(n\) is the number of moles, \(C_P\) is the molar heat capacity at constant pressure, and \(\Delta T\) is the change in temperature. Solve for \(\Delta T\) to find the increase in temperature.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Thermodynamics
Thermodynamics is a branch of physics that deals with the relationships between heat and other forms of energy. At its core, it involves studying how energy is converted from one form to another and how it affects matter. It is governed by several fundamental laws, which include the first law, often referred to as the law of energy conservation. This law states that energy cannot be created or destroyed, only transferred or converted from one form to another.

The exercise in question touches upon the first law of thermodynamics by examining how the transfer of heat (\( q \)), to water results in an increase in temperature. It is essential to understand that the thermal energy transferred to the water will cause the water molecules to move more rapidly, leading to a rise in temperature. This temperature change (\( \triangle T \)) is an important concept in thermodynamics as it is a measurable physical quantity that can often be directly related to the quantity of energy transferred.
Deciphering Molar Heat Capacity
Molar heat capacity (\( C_P \)) is a significant concept within thermodynamics and is a measure of the amount of heat required to raise the temperature of one mole of a substance by one Kelvin at constant pressure. It is intrinsic to understanding how different materials respond to heat. Molar heat capacity is a specific property of a material and is critical for predicting temperature changes when energy is added or removed.

In the textbook exercise, the molar heat capacity of water is given, allowing for calculations involving heat transfer and temperature changes. Knowing the molar heat capacity is necessary for targeting the precise energy requirements for temperature adjustments in various applications, from industrial processes to everyday cooking. It's why we can predict that adding a certain amount of heat to a specific mass of water will result in a definable increase in temperature.
The Mechanics of Heat Transfer
Heat transfer refers to the process of thermal energy moving from a hotter object to a cooler one. It is a fundamental concept within thermodynamics and occurs in various ways, including conduction, convection, and radiation. In our daily lives, we witness heat transfer when a warm coffee cup cools down to room temperature or when the sun heats the surface of the Earth.

In the context of the exercised problem, heat transfer is explored quantitatively by applying a specific amount of heat to water and observing the resultant change in temperature. The heat transfer formula \( q = nC_P\triangle T \) connects the dots between the heat supplied (\( q \)), the substance's molar heat capacity (\( C_P \)), and the resultant temperature change (\( \triangleT \)). By manipulating the formula, we can solve for the unknown temperature change once the amount of heat added and the molar heat of the substance are known. This process not only illustrates how temperature change is directly related to heat transfer but also how substances react uniformly based on their individual molar heat capacities.

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Most popular questions from this chapter

The work done during the expansion of a gas from \(4 \mathrm{dm}^{3}\) to \(6 \mathrm{dm}^{3}\) against a constant external pressure of \(3 \mathrm{~atm}\) is \((1 \mathrm{~L} \mathrm{~atm}=101.32 \mathrm{~J})\) (a) \(-6]\) (b) \(-608 J\) (c) \(+304 \mathrm{~J}\) (d) \(-304 \mathrm{~J}\)

If enthalpy of an overall reaction \(X \rightarrow Y\) along one route is \(\Delta_{r} H\) and \(\Delta_{r} H_{1}, \Delta_{1} H_{2}, \Delta_{r} H_{3} \ldots\) representing enthalpies of reactions leading to same product \(y\) then \(\Delta_{r} H\) is (a) \(\Delta_{r} H=\Delta_{r} H_{1}+\Delta_{r} H_{2}+\Delta_{r} H_{3} \ldots\) (b) \(\Delta_{r} H=\Delta_{r} H_{1} \times \Delta_{r} H_{2} \times \Delta_{r} H_{3} \ldots\) (c) \(\Delta_{r} H=\Delta_{r} H_{1}+\Delta_{r} H_{2}-\Delta_{r} H_{3} \ldots\) (d) \(\Delta_{r} H=\frac{\Delta_{r} H_{1} \times \Delta_{r} H_{2} \times \Delta_{r} H_{3}}{2} \cdots\)

In a reaction \(P+Q \rightarrow R+S\), there is no change in entropy. Enthalpy change for the reaction \((\Delta H)\) is \(12 \mathrm{~kJ} \mathrm{~mol}^{-1}\). Under what conditions, reaction will have negative value of free energy change? (a) If \(\Delta H\) is positive. (b) If \(\Delta H\) is negative. (c) If \(\Delta H\) is \(24 \mathrm{~kJ} \mathrm{~mol}^{-1}\). (d) If temperature of reaction is high.

Match the following columns and mark the appropriate choice. \(\begin{array}{|l|l|l|l|} \hline {\text { Column I }} & & {\text { Column II }} \\ \hline \text { (A) } & \text { Exothermic } & \text { (i) } & \Delta H=0, \Delta E=0 \\ \hline \text { (B) } & \text { Spontaneous } & \text { (ii) } & \Delta G=0 \\ \hline \text { (C) } & \text { Cyclic process } & \text { (iii) } & \Delta H \text { is negative. } \\ \hline \text { (D) } & \text { Equilibrium } & \text { (iv) } & \Delta G \text { is negative. } \\ \hline \end{array}\) (a) \((\mathrm{A}) \rightarrow(\mathrm{ii}),(\mathrm{B}) \rightarrow(\mathrm{iii}),(\mathrm{C}) \rightarrow(\mathrm{i}),(\mathrm{D}) \rightarrow(\mathrm{iv})\) (b) \((\mathrm{A}) \rightarrow(\mathrm{iv}),(\mathrm{B}) \rightarrow(\mathrm{i}),(\mathrm{C}) \rightarrow(\mathrm{iii}),(\mathrm{D}) \rightarrow\) (ii) (c) (A) \(\rightarrow\) (i), (B) \(\rightarrow\) (ii), (C) \(\rightarrow\) (iv), (D) \(\rightarrow\) (iii) (d) \((A) \rightarrow(i i i),(B) \rightarrow(i v),(C) \rightarrow(i),(D) \rightarrow(i i)\)

The equilibrium constant for a reaction is \(10 .\) What will be the value of \(\Delta G^{\circ}\) at \(300 \mathrm{~K} ?\) (a) \(-5.74 \mathrm{~kJ}\) (b) \(-574 \mathrm{~kJ}\) (c) \(+11.48 \mathrm{~kJ}\) (d) \(+5.74 \mathrm{~kJ}\)

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