
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
148.5k+ 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 2021 July 25 Shift 1 Question Paper with Answer Key

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

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Chemical Properties of Hydrogen - Important Concepts for JEE Exam Preparation

Trending doubts
A boy wants to throw a ball from a point A so as to class 11 physics JEE_Main

Select incorrect statements A Zero acceleration of class 11 physics JEE_Main

Assertion On a rainy day it is difficult to drive a class 11 physics JEE_Main

JEE Main Response Sheet 2025 Released – Download Links, and Check Latest Updates

How Many Students Will Appear in JEE Main 2025?

JEE Main Answer Key 2025 OUT; Download Session 1 NTA Official PDF Here

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

JEE Main Participating Colleges 2024 - A Complete List of Top Colleges

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

A solid cube and a solid sphere of the same material class 11 physics JEE_Main

The graph shows the variation of displacement of a class 11 physics JEE_Main

JEE Advanced Study Plan for 2025: Tips, Timetable, and Strategy
