
$100\text{ g}$ water is heated from $30{}^\circ \text{ C}$ to$50{}^\circ \text{ C}$ . Ignoring the slight expansion of the water, the change in its internal energy is (specific heat of water is 4184$\text{J/kg K}$ )
(A) $4\cdot 2\text{ K J}$
(B) $8\cdot 4\text{ K J}$
(C) $\text{84 K J}$
(D) $2\cdot 1\text{ K J}$
Answer
219.6k+ views
Hint The formula for internal energy (or heat) is equal to:
$\text{Q}=\text{mc}\vartriangle \text{T}$
By substituting the given values in the above equation, the answer can be formed
Complete step by step solution
We know that,
$\begin{align}
& \text{c}=4184\text{ j/kg K} \\
& \text{m}=100\text{ gm} \\
& \text{ }=0\cdot 1\text{ kg} \\
\end{align}$
$\begin{align}
& \vartriangle \text{T}=50{}^\circ \text{ C}-30{}^\circ \text{ C} \\
& \text{ }=20{}^\circ \text{ C} \\
& \text{ }=323\cdot 15\text{ K}-303\cdot 15\text{ K} \\
& \text{ }=20\text{ K} \\
& \text{ }\left[ {}^\circ \text{C}+273\cdot 15=\text{K} \right] \\
\end{align}$
We know that change in internal energy is equal to heat dissipated and is given by:
$\text{Q}=\text{mc}\vartriangle \text{T}$
Substituting the above values in this equation, we get
$\begin{align}
& \text{Q}=0\cdot 1\times 4184\times 20 \\
& \text{ }=8368\text{ J} \\
\end{align}$
\[\begin{align}
& \text{Q}=8\cdot 368\text{ K J} \\
& \text{ }=8\cdot 4\text{ K J} \\
\end{align}\]
So correct option is (B)
Note The heat capacity measures the amount of heat which is important to raise the temperature of an object or system by one degree calcium. It is denoted by C.
$\text{C}=\frac{\text{Q}}{\text{m}\vartriangle \text{T}}$
Or $\text{Q}=\text{mc}\vartriangle \text{T}$
Heat capacity is an extensive property. It depends upon the temperature, pressure and volume of the system under consideration.
The heat transferred Q causes a temperature $\vartriangle \text{T}$. The amount of heat transfer also depends upon mass.
$\text{Q}=\text{mc}\vartriangle \text{T}$
By substituting the given values in the above equation, the answer can be formed
Complete step by step solution
We know that,
$\begin{align}
& \text{c}=4184\text{ j/kg K} \\
& \text{m}=100\text{ gm} \\
& \text{ }=0\cdot 1\text{ kg} \\
\end{align}$
$\begin{align}
& \vartriangle \text{T}=50{}^\circ \text{ C}-30{}^\circ \text{ C} \\
& \text{ }=20{}^\circ \text{ C} \\
& \text{ }=323\cdot 15\text{ K}-303\cdot 15\text{ K} \\
& \text{ }=20\text{ K} \\
& \text{ }\left[ {}^\circ \text{C}+273\cdot 15=\text{K} \right] \\
\end{align}$
We know that change in internal energy is equal to heat dissipated and is given by:
$\text{Q}=\text{mc}\vartriangle \text{T}$
Substituting the above values in this equation, we get
$\begin{align}
& \text{Q}=0\cdot 1\times 4184\times 20 \\
& \text{ }=8368\text{ J} \\
\end{align}$
\[\begin{align}
& \text{Q}=8\cdot 368\text{ K J} \\
& \text{ }=8\cdot 4\text{ K J} \\
\end{align}\]
So correct option is (B)
Note The heat capacity measures the amount of heat which is important to raise the temperature of an object or system by one degree calcium. It is denoted by C.
$\text{C}=\frac{\text{Q}}{\text{m}\vartriangle \text{T}}$
Or $\text{Q}=\text{mc}\vartriangle \text{T}$
Heat capacity is an extensive property. It depends upon the temperature, pressure and volume of the system under consideration.
The heat transferred Q causes a temperature $\vartriangle \text{T}$. The amount of heat transfer also depends upon mass.
Recently Updated Pages
Electricity and Magnetism Explained: Key Concepts & Applications

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

States of Matter Chapter For JEE Main Chemistry

Trending doubts
Understanding Uniform Acceleration in Physics

Understanding Atomic Structure for Beginners

Understanding Centrifugal Force in Physics

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Degree of Dissociation: Meaning, Formula, Calculation & Uses

Understanding Average and RMS Value in Electrical Circuits

Other Pages
NCERT Solutions for Class 11 Physics Chapter 6 System Of Particles And Rotational Motion 2025-26

Understanding Entropy Changes in Different Processes

Common Ion Effect: Concept, Applications, and Problem-Solving

Understanding Excess Pressure Inside a Liquid Drop

NCERT Solutions For Class 11 Physics Chapter 12 Kinetic Theory - 2025-26

Motion In A Plane Class 11 Physics Chapter 3 CBSE Notes - 2025-26

