The thermal capacity of 40 g of aluminium \[\left( {\text{specific heat }= 0.2\,cal/g{}^0C} \right)\] is?
A. \[40cal/{}^0C\]
B. \[160cal/{}^0C\]
C. \[200cal/{}^0C\]
D. \[8cal/{}^0C\]
Answer
249.6k+ views
Hint: The thermal capacity of a body is defined as the heat required to raise the temperature of the given body by one degree. On the other hand we can also express the heat capacity formula as the product of mass, specific heat, and change in temperature. Change in thermal energy can be calculated by the formula:Change in thermal energy = mass $\times$ specific heat capacity $\times$ change in temperature. Here the change in temperature is calculated by subtracting the initial temperature from the final temperature.
Formula used:
Specific heat capacity is given as:
\[Q = mc\Delta T\]
Where Q is the heat energy, m is the mass, c is the specific heat capacity and \[\Delta T\] is the change in temperature.
Complete step by step solution:
Given mass of aluminium, m= 40 g
\[\left( {\text{specific heat }= 0.2\,cal/g{}^0C} \right)\]
As we know that thermal capacity,
\[Q = m \times c \times \Delta t\]
Here we see that \[Q \propto c\].
Substituting the given values, we get
Thermal capacity, \[Q = 40 \times 0.2 \times 1\]
\[\therefore {\rm{ Q = 8\,cal/}}{}^0{\rm{C}}\]
Therefore, the thermal capacity of 40 g of aluminium \[\left( {\text{specific heat }= 0.2\,cal/g{}^0C} \right)\] is \[{\rm{8cal/}}{}^0{\rm{C}}\].
Hence option D is the correct answer
Note: Heat capacity is defined as the heat required to increase the temperature of an object by one degree. Heat gain as well as heat loss results in changes in temperature, state and performance of the work. Heat is also known to be a transfer of energy. The heat capacity is usually expressed in joules or in calories and temperature is expressed in Celsius or Kelvin. Any heat capacity problem can be calculating the heat capacity, mass or temperature difference of any given substance.
Formula used:
Specific heat capacity is given as:
\[Q = mc\Delta T\]
Where Q is the heat energy, m is the mass, c is the specific heat capacity and \[\Delta T\] is the change in temperature.
Complete step by step solution:
Given mass of aluminium, m= 40 g
\[\left( {\text{specific heat }= 0.2\,cal/g{}^0C} \right)\]
As we know that thermal capacity,
\[Q = m \times c \times \Delta t\]
Here we see that \[Q \propto c\].
Substituting the given values, we get
Thermal capacity, \[Q = 40 \times 0.2 \times 1\]
\[\therefore {\rm{ Q = 8\,cal/}}{}^0{\rm{C}}\]
Therefore, the thermal capacity of 40 g of aluminium \[\left( {\text{specific heat }= 0.2\,cal/g{}^0C} \right)\] is \[{\rm{8cal/}}{}^0{\rm{C}}\].
Hence option D is the correct answer
Note: Heat capacity is defined as the heat required to increase the temperature of an object by one degree. Heat gain as well as heat loss results in changes in temperature, state and performance of the work. Heat is also known to be a transfer of energy. The heat capacity is usually expressed in joules or in calories and temperature is expressed in Celsius or Kelvin. Any heat capacity problem can be calculating the heat capacity, mass or temperature difference of any given substance.
Recently Updated Pages
JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Instantaneous Velocity Explained: Formula, Examples & Graphs

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

Hybridisation in Chemistry – Concept, Types & Applications

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

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

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

NCERT Solutions For Class 11 Physics Chapter 1 Units And Measurements - 2025-26

CBSE Notes Class 11 Physics Chapter 4 - Laws of Motion - 2025-26

