LEVELJEE Main

Animated Solution for Physics - Thermodynamics: A rigid container with thermally insulated walls contains a coil of resistance $100 Ω$ , carrying current $1 A$ . Change in internal energy after $5 min$ will be

Q: Why do we use the formula $i^2Rt$ instead of $V^2/R \times t$ for the heat generated?

We use $i^2Rt$ because the problem explicitly provides the current ($i = 1\text{ A}$) and the resistance ($R = 100\ \Omega$). We do not know the voltage across the coil.

Q: What happens to the temperature of the gas inside the container?

Since the internal energy of the gas increases (by $30\text{ kJ}$), and internal energy is directly proportional to temperature for an ideal gas, the temperature of the gas will increase.

Select Answer:

Visualized Solution

$Visualizing the Setup$

$R = 100 Ω$
$i = 1 A$
$t = 5 min$

$First Law of Thermodynamics$

$Δ Q = Δ U + W$

$Work Done$

$Rigid container ⟹ Δ V = 0$
$W = P Δ V = 0$

$Joule Heating$

$Heat supplied by coil:$
$H = i^{2} R t$

$Energy Balance$

$Δ Q = H$
$Δ U = i^{2} R t$

$Substitution$

$t = 5 min = 300 s$
$Δ U = (1)^{2} \times 100 \times 300$

$Calculation$

$Δ U = 1 \times 100 \times 300$
$Δ U = 30000 J$

$Final Answer$

$Δ U = 30 kJ$

00:00 / 00:00

The Sigma Insight: First Law of Thermodynamics

Solution Diagram

Analyzing the Setup

Imagine you are looking at a rigid, thermally insulated container. Inside this sealed vault, there is a heating coil with a resistance of

100 Ω

, and a steady current of

1 A

is flowing through it.

The problem asks for the change in internal energy of the system after

5 min

. To crack this, we need to bridge two beautiful concepts: Joule Heating from Current Electricity and the First Law of Thermodynamics.

The Master Equation

Let's start with the First Law of Thermodynamics, which is essentially the law of conservation of energy for heat systems:

Δ Q = Δ U + W

Here is the catch: the container is rigid. This means its volume is absolutely fixed (

Δ V = 0

). If the volume cannot expand or compress, the gas cannot do any work on its surroundings.

W = P Δ V = 0

Because the walls are thermally insulated, no heat can enter or escape from the outside. However, the heating coil is inside the container! The electrical energy dissipated by the coil acts as the heat source for the gas.

Therefore, the entire electrical heat generated goes directly into increasing the internal energy of the system:

Δ U = Δ Q = i^{2} R t

Final Calculation

Now, it is just a matter of plugging in the numbers. But watch out for the units! The time is given in minutes, and we must convert it to seconds to stay in the standard SI system.

i = 1 A

R = 100 Ω

t = 5 min = 5 \times 60 s = 300 s

Substituting these into our equation:

Δ U = (1)^{2} \times 100 \times 300

Δ U = 30000 J

To make it neater, we convert Joules to kilo-Joules:

Δ U = 30 kJ

And there we have it! The internal energy of the system increases by exactly

30 kJ

Common Traps & FAQs

Why is the work done by the gas zero?

The container is described as 'rigid', which means its volume is completely fixed. Since work done by a gas is given by

W = P Δ V

, and

Δ V = 0

, the work done is exactly zero.

Doesn't 'thermally insulated' mean that the heat transfer

Δ Q

is zero?

It means no heat can enter or leave from the *outside* environment. However, the heating coil is located *inside* the container, so it acts as an internal heat source adding energy directly to the gas.

Why do we use the formula

i^{2} R t

instead of

V^{2} / R \times t

for the heat generated?

We use

i^{2} R t

because the problem explicitly provides the current (

i = 1 A

) and the resistance (

R = 100 Ω

). We do not know the voltage across the coil.

What happens to the temperature of the gas inside the container?

Since the internal energy of the gas increases (by

30 kJ

), and internal energy is directly proportional to temperature for an ideal gas, the temperature of the gas will increase.

Why do we need to convert the time from minutes to seconds?

To obtain the energy in Joules (the standard SI unit), all input values must be in their respective SI units. Time must be in seconds, so

5 min

becomes

300 s

Similar Questions

LEVELJEE Main

An insulated container of gas has two chambers separated by an insulating partition. One of the chambers has volume $V_{1}$ and contains ideal gas at pressure $p_{1}$ and temperature $T_{1}$ . The other chamber has volume $V_{2}$ and contains ideal gas at pressure $p_{2}$ and temperature $T_{2}$ . If the partition is removed without doing any work on the gas, the final equilibrium temperature of the gas in the container will be

(A)

\frac{T _{1} T _{2} ( p _{1} V _{1} + p _{2} V _{2} )}{p _{1} V _{1} T _{2} + p _{2} V _{2} T _{1}}

(B)

\frac{p _{1} V _{1} T _{1} + p _{2} V _{2} T _{2}}{p _{1} V _{1} + p _{2} V _{2}}

(C)

\frac{p _{1} V _{1} T _{2} + p _{2} V _{2} T _{1}}{p _{1} V _{1} + p _{2} V _{2}}

(D)

\frac{T _{1} T _{2} ( p _{1} V _{1} + p _{2} V _{2} )}{p _{1} V _{1} T _{1} + p _{2} V _{2} T _{2}}

LEVELJEE Main

A thermally insulated vessel contains an ideal gas of molecular mass $M$ and ratio of specific heats $γ$ . It is moving with speed $v$ and its suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by

(A)

\frac{( γ - 1 )}{2 γ R} M v^{2} K

(B)

\frac{γ M v ^{2}}{2 R} K

(C)

\frac{( γ - 1 )}{2 R} M v^{2} K

(D)

\frac{( γ - 1 )}{2 ( γ + 1 ) R} M v^{2} K

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An electric appliance supplies $6000 J/min$ heat to the system. If the system delivers a power of $90 W$ . How long it would take to increase the internal energy by $2.5 \times 1 0^{3} J$ ?

(A)

2.5 \times 1 0^{2} s

(B)

4.1 \times 1 0^{1} s

(C)

2.4 \times 1 0^{3} s

(D)

2.5 \times 1 0^{1} s

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A container with insulating walls is divided into two equal parts by a partition fitted with a valve. One part is filled with an ideal gas at a pressure $p$ and temperature $T$ , whereas the other part is completely evacuated. If the valve is suddenly opened, the pressure and temperature of the gas will be

(A)

\frac{p}{2}, \frac{T}{2}

(B)

\frac{p}{2}, T

(C)

p, T

(D)

p, \frac{T}{2}

JEE Advanced 2026

LEVELJEE Advanced

As shown in the figure, an insulated container is fitted with a thermally conducting but immovable partition ( $P_{1}$ ) and a freely movable but thermally insulated piston ( $P_{2}$ ). The partition $P_{1}$ with thermal conductivity $K$ , cross sectional area $A$ and width $x$ divides the container into two sections, $S_{1}$ and $S_{2}$ , each containing one mole of a monoatomic gas. The piston $P_{2}$ moves freely such that the gas in $S_{2}$ is always at the atmospheric pressure. Initially, the difference between the temperatures of $S_{1}$ and $S_{2}$ is $Δ T_{0}$ . The time it takes for the temperature difference to become $\frac{Δ T _{0}}{2}$ is $n x R / K A$ , where $R$ is the universal gas constant. The value of $n$ is: [Given: $ln 2 \approx 0.7$ ]

LEVELJEE Main

A container of volume $1 m^{3}$ is divided into two equal parts by a partition. One part has an ideal gas at $300 K$ and the other part is vacuum. The whole system is thermally isolated from the surroundings. When the partition is removed, the gas expands to occupy the whole volume. Its temperature will now be ......

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LEVELJEE Main

The internal energy change when a system goes from state $A$ to $B$ is $40 kJ/mol$ . If the system goes from $A$ to $B$ by a reversible path and returns to state $A$ by an irreversible path, what would be the net change in internal energy?

(A)

40 kJ

(B)

> 40 kJ

(C)

< 40 kJ

Comprehension Passage

In the figure a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of an ideal monoatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K. The heat capacities per mole of an ideal monoatomic gas are

C_{V} = \frac{3}{2} R, C_{p} = \frac{5}{2} R

, and those for an ideal diatomic gas are

C_{V} = \frac{5}{2} R, C_{p} = \frac{7}{2} R

Question 1:

Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be

Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be

An ideal gas in thermally insulated vessel at internal pressure = $P_{1}$ , volume = $V_{1}$ and absolute temperature = $T_{1}$ expands irrversibly against zero external pressure, as shown in the diagram. The final internal pressure, volume and absolute temperature of the gas are $P_{2}$ , $V_{2}$ and $T_{2}$ , respectively. For this expansion,

* Multiple Correct Options

(A)

q = 0

(B)

T_{2} = T_{1}

(C)

P_{2} V_{2} = P_{1} V_{1}

(D)

P_{2} V_{2}^{γ} = P_{1} V_{1}^{γ}

LEVELJEE Main

For an ideal gas

* Multiple Correct Options

(A)

the change in internal energy in a constant pressure process from temperature

T_{1}

T_{2}

is equal to

n C_{V} (T_{2} - T_{1})

, where

C_{V}

is the molar heat capacity at constant volume and

n

the number of moles of the gas

(B)

the change in internal energy of the gas and the work done by the gas are equal in magnitude in an adiabatic process

(C)

the internal energy does not change in an isothermal process

(D)

no heat is added or removed in an adiabatic process

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Law of Thermodynamics

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Work Done

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Joule Heating

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy Balance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substitution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer

The Sigma Insight: First Law of Thermodynamics

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } An electric appliance supplies 6000 J/min heat to the system. If the system delivers a power of 90 W. How long it would take to increase the internal energy by 2.5×103 J?

Comprehension Passage

Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be

Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be

For an ideal gas

$Visualizing the Setup$

$First Law of Thermodynamics$

$Work Done$

$Joule Heating$

$Energy Balance$

$Substitution$

$Calculation$

$Final Answer$

An electric appliance supplies $6000 J/min$ heat to the system. If the system delivers a power of $90 W$ . How long it would take to increase the internal energy by $2.5 \times 1 0^{3} J$ ?