
Calculate the amount of heat energy required to raise the temperature of $100{\text{g}}$ of copper from $20^\circ {\text{C}}$ to $70^\circ {\text{C}}$. Specific heat capacity of copper $ = 390{\text{Jk}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$.
A) $1950{\text{J}}$
B) $3900{\text{J}}$
C) $390{\text{J}}$
D) ${\text{None of the above}}$
Answer
144.6k+ views
Hint: An increase in temperature requires heat to be supplied. Now the amount of this heat will be proportional to the mass of the copper and the temperature difference between the current temperature of the copper and the temperature we desire. The specific heat capacity of copper will serve as the proportionality constant.
Formula used:
The amount of heat required is given by, $Q = mc\Delta T$ where $m$ is the mass of the sample, $c$ is the specific heat capacity of the material of the sample and $\Delta T$ is the change in temperature.
Complete step by step answer:
Step 1: List the parameters involved in the problem at hand.
The mass of the sample of copper is given to be $m = 100{\text{g}}$ .
The specific heat capacity of copper is given to be $c = 390{\text{Jk}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$ .
Since the temperature has to be raised from $20^\circ {\text{C}}$ to $70^\circ {\text{C}}$ , the change in temperature will be $\Delta T = 70 - 20 = 50^\circ {\text{C}}$ .
Step 2: Express the relation for the heat required.
The amount of heat required to bring about the necessary change in temperature can be expressed as $Q = mc\Delta T$ --------- (1)
Substituting for $m = 0.1{\text{kg}}$ , $c = 390{\text{Jk}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$ and $\Delta T = 50{\text{K}}$ in equation (1) we get, $Q = 0.1 \times 390 \times 50 = 1950{\text{J}}$
$\therefore $ the heat required to bring about the necessary change in temperature is obtained to be $Q = 1950{\text{J}}$ .
So the correct option is A.
Note: While substituting values of different physical quantities in any equation make sure that all the quantities are expressed in their respective S.I units. If not, then the necessary conversion of units must be done. Here the mass of the sample was expressed in the unit of grams so we expressed it in the S.I unit of kilogram as $m = 0.1{\text{kg}}$ before substituting in equation (1). However, the change in the temperature is basically a difference between two temperatures and its value in the Celsius scale and the Kelvin scale are the same, so we do not make a conversion of units for the change in temperature $\Delta T$ .
Formula used:
The amount of heat required is given by, $Q = mc\Delta T$ where $m$ is the mass of the sample, $c$ is the specific heat capacity of the material of the sample and $\Delta T$ is the change in temperature.
Complete step by step answer:
Step 1: List the parameters involved in the problem at hand.
The mass of the sample of copper is given to be $m = 100{\text{g}}$ .
The specific heat capacity of copper is given to be $c = 390{\text{Jk}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$ .
Since the temperature has to be raised from $20^\circ {\text{C}}$ to $70^\circ {\text{C}}$ , the change in temperature will be $\Delta T = 70 - 20 = 50^\circ {\text{C}}$ .
Step 2: Express the relation for the heat required.
The amount of heat required to bring about the necessary change in temperature can be expressed as $Q = mc\Delta T$ --------- (1)
Substituting for $m = 0.1{\text{kg}}$ , $c = 390{\text{Jk}}{{\text{g}}^{ - 1}}{{\text{K}}^{ - 1}}$ and $\Delta T = 50{\text{K}}$ in equation (1) we get, $Q = 0.1 \times 390 \times 50 = 1950{\text{J}}$
$\therefore $ the heat required to bring about the necessary change in temperature is obtained to be $Q = 1950{\text{J}}$ .
So the correct option is A.
Note: While substituting values of different physical quantities in any equation make sure that all the quantities are expressed in their respective S.I units. If not, then the necessary conversion of units must be done. Here the mass of the sample was expressed in the unit of grams so we expressed it in the S.I unit of kilogram as $m = 0.1{\text{kg}}$ before substituting in equation (1). However, the change in the temperature is basically a difference between two temperatures and its value in the Celsius scale and the Kelvin scale are the same, so we do not make a conversion of units for the change in temperature $\Delta T$ .
Recently Updated Pages
Difference Between Rows and Columns: JEE Main 2024

Difference Between Natural and Whole Numbers: JEE Main 2024

How to find Oxidation Number - Important Concepts for JEE

How Electromagnetic Waves are Formed - Important Concepts for JEE

Electrical Resistance - Important Concepts and Tips for JEE

Average Atomic Mass - Important Concepts and Tips for JEE

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Degree of Dissociation and Its Formula With Solved Example for JEE

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

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry
