Answer
Verified
113.7k+ 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
JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key
Geostationary Satellites and Geosynchronous Satellites for JEE
Complex Numbers - Important Concepts and Tips for JEE
JEE Main 2023 (February 1st Shift 2) Maths Question Paper with Answer Key
JEE Main 2022 (July 25th Shift 2) Physics Question Paper with Answer Key
Inertial and Non-Inertial Frame of Reference for JEE
Trending doubts
JEE Main 2025: Application Form (Out), Exam Dates (Released), Eligibility & More
JEE Main Chemistry Question Paper with Answer Keys and Solutions
Class 11 JEE Main Physics Mock Test 2025
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
JEE Main Login 2045: Step-by-Step Instructions and Details
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics
Other Pages
NCERT Solutions for Class 11 Physics Chapter 7 Gravitation
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids
Units and Measurements Class 11 Notes - CBSE Physics Chapter 1
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements
NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
NCERT Solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids