As a physicist, you put heat into a $500g$ solid sample at the rate of $10.0kJ{\min ^{ - 1}}$, while recording its temperature as a function of time. You plot your data and obtain the graph as shown in the figure. What is the specific heat of the solid state of the material.
(A). $2.66kJk{g^{ - 1}}^ \circ {C^{ - 1}}$
(B). \[4.99kJk{g^{ - 1}}^ \circ {C^{ - 1}}\]
(C). $1.33kJk{g^{ - 1}}^ \circ {C^{ - 1}}$
(D). None of these

Answer
301.2k+ views
Hint: We know that a solid changes into a liquid at a temperature equal to melting point of the solid. Similarly a liquid changes into a gaseous state at a temperature equals to boiling point of liquid. Temperature remains unaffected as the state of the entire substance has changed. Thus a thermometer fails to detect the change of state as there is neither rise nor fall in temperature duriv the change of state. That is why heat energy spent in change of state is called Latent Heat.
$L = \dfrac{Q}{m}$
Here, L is the latent heat, Q is the heat absorbed and m is the mass of the body.
Latent heat is measured in $Jk{g^{ - 1}}$ or $\operatorname{Cal} {\operatorname{g} ^{ - 1}}$. For example Latent heat of Fusion of ice \[ = 80{\text{ }}\operatorname{Ca} {\lg ^{ - 1}} = 3.33 \times {10^J}Jk{g^{ - 1}}\]
Complete step by step answer:
Now from the graph it is clear that within the time period,Time $\left( t \right) = 1\min $ to Time $\left( t \right) = 3\min $
There is no change/rise in the temperature of the substance. Therefore the solid melts in this time period.
Now the Latent heat we have is
$\implies$ $L = \dfrac{Q}{m} = \dfrac{{HT}}{m}{\text{ - - - - - - - }}\left( 1 \right)$
Here $H$ is rate of change of heat
$T$ is the time period for heat change.
So now here we have from the question
$H = 10.0kJ{\min ^{ - 1}}$
And $T = 2$
$m = 0.5$
Now putting all the above values in the equation $\left( 1 \right)$ so we have.
$\implies$ $L = \dfrac{{10 \times 2}}{{0.5}} = 40kJk{g^{ - 1}}$
And also we have the formula for specific heat is
$\implies$ $\Delta Q = ms\left( {\Delta T} \right)$
Here $\Delta Q$ is the small amount of heat energy required to raise the temperature of certain mass $\left( m \right)$ of a substance through a small range of temperature $\left( {\Delta T} \right)$.
Here $'S'$ in the formula is called a Constant of proportionality and is called specific heat capacity or simply specific heat of the material of the substance. $Q = mS\Delta T$
So now $S = \dfrac{Q}{{m \times \Delta T}}$
Now $S = \dfrac{{10 \times 1}}{{0.5 \times 15}} = 1.33kJk{g^{ - 1}}^ \circ {C^{ - 1}}$
Hence option ‘C’ is correct.
Note: Molar specific heat or molar heat capacity:
It is defined as the amount of heat required to raise the temperature of one gram mole of the substance through a unit degree.
It is represented by ‘C’ .
Now by the definition.
One mole of any substance is Quantity of substance whose mass in gram is numerically equals to the molecular mass $\left( M \right)$
$C = Ms$
To calculate the no of moles in a given $\left( {m\operatorname{gm} } \right)$ of the substance, we divide $'m'$ buy the molecular mass $\left( M \right)$
$n = \dfrac{m}{M}$ or $m = nM$
Now from the definition of specific heat
$S = \dfrac{{\Delta Q}}{{m\Delta T}} = \dfrac{{\Delta Q}}{{nM\left( {\Delta T} \right)}}$
$MS = \dfrac{1}{n}\left( {\dfrac{{\Delta Q}}{{\Delta T}}} \right)$
$C = \dfrac{1}{n}\left( {\dfrac{{\Delta Q}}{{\Delta T}}} \right)$
The above is the expression for the molar specific heat of the substance.
$L = \dfrac{Q}{m}$
Here, L is the latent heat, Q is the heat absorbed and m is the mass of the body.
Latent heat is measured in $Jk{g^{ - 1}}$ or $\operatorname{Cal} {\operatorname{g} ^{ - 1}}$. For example Latent heat of Fusion of ice \[ = 80{\text{ }}\operatorname{Ca} {\lg ^{ - 1}} = 3.33 \times {10^J}Jk{g^{ - 1}}\]
Complete step by step answer:
Now from the graph it is clear that within the time period,Time $\left( t \right) = 1\min $ to Time $\left( t \right) = 3\min $
There is no change/rise in the temperature of the substance. Therefore the solid melts in this time period.
Now the Latent heat we have is
$\implies$ $L = \dfrac{Q}{m} = \dfrac{{HT}}{m}{\text{ - - - - - - - }}\left( 1 \right)$
Here $H$ is rate of change of heat
$T$ is the time period for heat change.
So now here we have from the question
$H = 10.0kJ{\min ^{ - 1}}$
And $T = 2$
$m = 0.5$
Now putting all the above values in the equation $\left( 1 \right)$ so we have.
$\implies$ $L = \dfrac{{10 \times 2}}{{0.5}} = 40kJk{g^{ - 1}}$
And also we have the formula for specific heat is
$\implies$ $\Delta Q = ms\left( {\Delta T} \right)$
Here $\Delta Q$ is the small amount of heat energy required to raise the temperature of certain mass $\left( m \right)$ of a substance through a small range of temperature $\left( {\Delta T} \right)$.
Here $'S'$ in the formula is called a Constant of proportionality and is called specific heat capacity or simply specific heat of the material of the substance. $Q = mS\Delta T$
So now $S = \dfrac{Q}{{m \times \Delta T}}$
Now $S = \dfrac{{10 \times 1}}{{0.5 \times 15}} = 1.33kJk{g^{ - 1}}^ \circ {C^{ - 1}}$
Hence option ‘C’ is correct.
Note: Molar specific heat or molar heat capacity:
It is defined as the amount of heat required to raise the temperature of one gram mole of the substance through a unit degree.
It is represented by ‘C’ .
Now by the definition.
One mole of any substance is Quantity of substance whose mass in gram is numerically equals to the molecular mass $\left( M \right)$
$C = Ms$
To calculate the no of moles in a given $\left( {m\operatorname{gm} } \right)$ of the substance, we divide $'m'$ buy the molecular mass $\left( M \right)$
$n = \dfrac{m}{M}$ or $m = nM$
Now from the definition of specific heat
$S = \dfrac{{\Delta Q}}{{m\Delta T}} = \dfrac{{\Delta Q}}{{nM\left( {\Delta T} \right)}}$
$MS = \dfrac{1}{n}\left( {\dfrac{{\Delta Q}}{{\Delta T}}} \right)$
$C = \dfrac{1}{n}\left( {\dfrac{{\Delta Q}}{{\Delta T}}} \right)$
The above is the expression for the molar specific heat of the substance.
Recently Updated Pages
Four persons A B C and D initially at the corners of class 11 physics JEE_Main

What is the difference between Conduction and conv class 11 physics JEE_Main

Moment of inertia of solid sphere about its diameter class 11 physics JEE_Main

If a piece of ice floating on the surface of water class 11 physics JEE_Main

At what temperature speed of sound in air will be doubled class 11 physics JEE_Main

A closed organ pipe and an open organ pipe are tuned class 11 physics JEE_Main

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

Understanding Atomic Structure for Beginners

Understanding Uniform Acceleration in Physics

Other Pages
CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2026-27 Free PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 2 Motion In A Straight Line - 2026-27 Free PDF Download (Login Required)

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2026-27 Free PDF Download (Sign-in Required)

CBSE Notes Class 11 Physics Chapter 2 - Motion in a Straight Line - 2026-27 PDF Download (Login Required)

NCERT Solutions For Class 11 Physics Chapter 3 Motion In A Plane - 2026-27 Free PDF Download (Sign-in Required)

