Heat

Heat is a fundamental form of energy that flows from one body to another due to a temperature difference. It plays a crucial role in everyday life, from cooking food and warming homes to powering engines and generating electricity. Heat energy is responsible for various physical processes, such as the melting of ice, boiling of water, and even the weather patterns we experience. In real life, understanding heat is essential for designing efficient heating and cooling systems, managing energy consumption, and ensuring the comfort and safety of living environments. For instance, insulating homes helps retain heat during winter, reducing energy bills and enhancing comfort. Thus, heat is not just a physical concept but a vital aspect of our daily lives, impacting everything from technology to climate control.

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This Story also Contains

1. Heat
2. Solved Examples Based on Heat
3. Summary

Heat

The form of energy which is exchanged among various bodies or systems on account of temperature difference is defined as heat. So, we can say that the driving potential for the heat energy is the temperature difference.

The temperature of a body can be changed by giving heat (temperature rises) or by removing heat (temperature falls) from the body. The amount of heat (Q) given to a body depends upon its mass (m), the change in its temperature and the nature of the material (C) i.e., $\mathrm{Q}=\mathrm{mC} \boldsymbol{\Delta} \theta$; where C = specific heat of material which depends on the material.

There are various units of heat like Joule(J), erg, calorie(cal) etc. Heat is a scalar quantity. The calorie (cal) is defined as the amount of heat required to raise the temperature of 1 gram of water by 1^oC.

1 cal = 4.186 J

There are basically two types of specific heats

Heat Transfer Occurs by Three Mechanisms

Conduction: It is the method in which the transfer of heat takes place between atoms and molecules in direct contact.

Convection: It is the method in which the transfer of heat happens by the movement of the heated substance.

Radiation: It is the method in which the transfer of heat takes place by electromagnetic waves.

Gram-Specific Heat

It is defined as the amount of heat energy required to raise the temperature of the unit mass of a body through 1°C (or K) is called gram-specific heat of the material of the body. Actually, it depends on the mass of the body which is in Gram.

If Q heat changes the temperature of mass m by $\Delta \theta$ then specific heat is given as $c=\frac{Q}{m \Delta \theta}$

Based on this equation we can calculate the unit and dimension of this

$\begin{aligned} & \text { SI unit is }=\frac{\text { Joule }}{k g-K} \\ & \text { Dimension is - }\left[L^2 T^{-2} \boldsymbol{\theta}^{-1}\right]\end{aligned}$

Molar Specific Heat

Molar specific heat of a substance 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.

It can be written as $C=M \frac{Q}{m \Delta \theta}=\frac{1}{\mu} \frac{Q}{\Delta \theta}$

Here,
$Q=$ Heat supplied, $M=$ Molecular mass, $m=$ Actual mass of the substance, $\Delta \theta=$ Temperature difference

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Solved Examples Based on Heat

Example 1: A body of mass 1 kg absorbs 4k cal of heat when the temperature rises from 30⁰ c to 50⁰ c, then specific heat capacity (in cal/g^oc) is

1)0.1

2)0.2

3)0.3

4)0.4

Solution:

Specific Heat

Amount of heat(Q) required to raise the temperature of unit mass through 1^oC.

wherein

$\begin{aligned} & \text { Units }- \text { Calories } / \mathrm{gm} \times{ }^{\circ} \mathrm{C} \\ & \quad \Delta Q=\mathrm{m} \mathrm{s} \Delta T \\ & \Rightarrow 4 \times 1000 \mathrm{cal}=1000 \times \mathrm{s} \times 20 \\ & s=0.2 \mathrm{cal} / \mathrm{g}^{\circ} \mathrm{C}\end{aligned}$

Hence, the answer is the option (2).

Example 2: Time (in seconds) taken by an 836 W heater to heat one litre of water from 10°C to 40°C is

1) 50

2) 100

3) 150

4) 200

Solution:

Specific Heat
$
C=\frac{Q}{m \cdot \Delta \theta}
$

wherein
$C=$ specific heat
$\Delta \theta=$ Change in temperature
$\mathrm{m}=$ Amount of mass
Amount of heat required

$
\begin{aligned}
& Q=m C \Delta T m C \Delta T=P t \Rightarrow 1 K g \times 4200 \times 30=P t \\
& t=\frac{126000}{836}=150 \mathrm{sec}
\end{aligned}
$

Hence, the answer is the option (3).

Example 3: A massless spring $(k=800 \mathrm{~N} / \mathrm{m})$, attached to a mass $(500 \mathrm{~g})$ is completely immersed in 1 kg of water. The spring stretched by 2 cm and released so that it started vibrating. What would be the order of magnitude of the change in the temperature of water when the vibrations stop completely? ( Assume that the water container and spring receive negligible heat and specific heat of mass $=400 \mathrm{~J} / \mathrm{kg} \mathrm{K}$, specific heat (in K) of water $4184 \mathrm{~J} / \mathrm{kg} \mathrm{K}$ )

1) 0.0001

2) 0.00003

3) 0.1

4) 0.001

Solution:

$\begin{aligned} & C=\frac{Q}{m \cdot \Delta \theta} \\ & \text { wherein } \\ & \mathrm{C}=\text { specific heat } \\ & \Delta \theta=\text { Change in temperature } \\ & \mathrm{m}=\text { Amount of mass } \\ & \text { Given } \\ & K=800 \mathrm{~N} / \mathrm{m} \\ & m=0.5 \mathrm{~kg} \\ & \frac{1}{2} K A^2=m_1 \mathrm{~s}_1 \Delta T+m_2 s_2 \Delta T \\ & \Rightarrow \frac{1}{2} \times 800 \times\left(\frac{2}{100}\right)^2=\frac{1}{2} \times 400 \times \Delta T+4184 \Delta T \\ & \Rightarrow \Delta T=\frac{400 \times\left(\frac{2}{100}\right)^2}{200+4184}=3.64 \times 10^{-5} \mathrm{~K}\end{aligned}$

Hence, the answer is the option (2).

Example 4: Molar heat capacity for a gas at constant pressure and constant volume is given as $C_p$ and $C_2$ respectively, then

1) $C_p=C_v$
2) $C_p<C_v$
3) $C_p>C_v$

4) can't be predicted

Solution:

Molar Specific Heat

$C=\frac{M Q}{m \cdot \Delta \theta}=\frac{1}{\mu} \frac{Q}{\Delta \theta}$

wherein

$\begin{aligned} & \frac{M}{m}=\mu \\ & \text { Unit }- \text { Calorie } / \text { mole } \times{ }^o \mathrm{C} \\ & C_p=C_v+R\end{aligned}$

Hence, the answer is the option (3).

Example 5: The density of ice is x g/cm³ and that of water is y g/cm³, when m gram of ice melts, then the change in volume is

1) $m(y-x)$
2) $\frac{y-x}{m}$
3) $m y(y-x)$
4) $\frac{m}{y}-\frac{m}{x}$

Solution:

After m gm of ice melts, the m/x volume of ice reduces & m/y volume of water increases

The volume of ice > volume of water

$\begin{aligned} \Delta V & =V_w-V_i \\ \Delta V & =\frac{m}{y}-\frac{m}{x} \\ \Delta V & =m\left[\frac{x-y}{x y}\right]\end{aligned}$

Hence, the answer is the option (4).

Summary

Heat is the transfer of energy from an object at a higher temperature to one at a lower temperature. It has vital applications in our lives, ranging from weathering to the various styles of cooking applied to food. In all, there are three main ways for heat transfer: through conduction, convection, and radiation. When one understands the methods, it allows for realizing how heat works in differing situations and how it can be effectively controlled.