A gas based geyser heats water flowing at the rate of 5.0 litres per minute from 27 to 87. The rate of consumption of the gas is ____ g/s. (Take heat of combustion of gas = 5.0 x 104 J/g, specific heat capacity of water = 4200 J/kg.)

Water flow rate = 5 l/min = (5/60) kg/s = (1/12) kg/s. The power required to heat the water is: Let the rate of consumption of gas be x g/s. Heat generated by the gas per second must equal the power required to heat the water: Equate the thermal power absorbed by fluid (dm/dt cdot c cdot Delta T) with the power output of fuel (dmfuel/dt cdot Lcombustion). Class 11 Physics: Thermal Properties of Matter

A gas based geyser heats water flowing at the rate of 5.0 litres per minute from 27 to 87. The rate of consumption of the gas is ____ g/s. (Take heat of combustion of gas = 5.0 x 104 J/g, specific hea

Water flow rate = 5 l/min = (5/60) kg/s = (1/12) kg/s. The power required to heat the water is: Let the rate of consumption of gas be x g/s. Heat generated by the gas per second must equal the power required to heat the water: Equate the thermal power absorbed by fluid (dm/dt cdot c cdot Delta T) with the power output of fuel (dmfuel/dt cdot Lcombustion). Class 11 Physics: Thermal Properties of Matter

A gas based geyser heats water flowing at the rate of 5.0 litres per minute from 27 to 87. The rate of consumption of the gas is ____ g/s. (Take heat of combustion of gas = 5.0 x 104 J/g, specific heat capacity of water = 4200 J/kg.)

Water flow rate = 5 l/min = (5/60) kg/s = (1/12) kg/s. The power required to heat the water is: Let the rate of consumption of gas be x g/s. Heat generated by the gas per second must equal the power required to heat the water: Equate the thermal power absorbed by fluid (dm/dt cdot c cdot Delta T) with the power output of fuel (dmfuel/dt cdot Lcombustion). Class 11 Physics: Thermal Properties of Matter

A gas based geyser heats water flowing at the rate of 5.0 litres per minute from 27 to 87. The rate of consumption of the gas is ____ g/s. (Take heat of combustion of gas = 5.0 x 104 J/g, specific hea

Water flow rate = 5 l/min = (5/60) kg/s = (1/12) kg/s. The power required to heat the water is: Let the rate of consumption of gas be x g/s. Heat generated by the gas per second must equal the power required to heat the water: Equate the thermal power absorbed by fluid (dm/dt cdot c cdot Delta T) with the power output of fuel (dmfuel/dt cdot Lcombustion). Class 11 Physics: Thermal Properties of Matter

A gas based geyser heats water flowing at the rate of 5.0 litres per minute from 27 to 87. The rate of consumption of…

Year	2026	2025	2024	Total
Questions	5	7	1	13

More Thermal Properties of Matter Previous-Year Questions

Q40 jee_main_2026_21_jan_morning Thermal Expansion

An aluminium and steel rods having same lengths and cross-sections are joined to make total length of 120 cm at 30°C. The coefficient of linear expansion of aluminium and steel are

24 \times 10⁻⁶/°C

and

1.2 \times 10⁻⁵/°C

, respectively. The length of this composite rod when its temperature is raised to 100°C, is ____ cm.

A. 120.20
B. 120.15
C. 120.03
D. 120.06

Solution

Related Formula

Δ l = l₀ α Δ T

lfinal = l₁ + l₂ + Δ l₁ + Δ l₂

Core Logic

Let the original lengths be $l0A$ $l_{0A}$ (aluminium) and $l0S$ $l_{0S}$ (steel). They have same length, so $l0A = l0S = 60 cm$ $l_{0A} = l_{0S} = 60\text{ cm}$. Change in temperature $Δ T = 100°C - 30°C = 70°C$ .

Expansion of Aluminium: $Δ lA = l₀ αA Δ T$ $\Delta l_A = l_0 \alpha_A \Delta T$ Expansion of Steel: $Δ lS = l₀ αS Δ T$ $\Delta l_S = l_0 \alpha_S \Delta T$

Step 1: Calculate Final Length

lfinal = l₀(1 + αA Δ T) + l₀(1 + αS Δ T)

lfinal = l₀ [2 + (αA + αS)Δ T]

lfinal = 60 [ 2 + (24 \times 10⁻⁶ + 12 \times 10⁻⁶) \times 70 ]

lfinal = 60 [ 2 + (36 \times 10⁻⁶) \times 70 ]

lfinal = 60 [2 + 0.00252] = 120 + 60(0.00252) = 120 + 0.1512 = 120.15 cm

Thermal Expansion diagram for Q40 - JEE Main 2026 Morning

Pattern Recognition

For composite rods connected end-to-end, expansions simply add up. Convert $1.2 \times 10⁻⁵$ $1.2 \times 10^{-5}$ to $12 \times 10⁻⁶$ $12 \times 10^{-6}$ immediately for easy mental addition.

Chapter Mix

Class 11 Physics: Thermal Properties of Matter

Q38 jee_main_2026_22_january_morning Thermal Conduction in Composite Rods

Rods x and y of equal dimensions but of different materials are joined as shown in figure. Temperatures of end points A and F are maintained at

100circC

and

40circC

respectively. Given the thermal conductivity of rod x is three times of that of rod y, the temperature at junction points B and E are (close to):

Thermal Properties of Matter diagram for Q38 - JEE Main 2026 January Morning — Composite rods x and y joined end-to-end with temperature endpoints.

A. $89circC$ $89^{circ}\text{C}$ and $73circC$ $73^{circ}\text{C}$ respectively
B. $80circC$ $80^{circ}\text{C}$ and $60circC$ $60^{circ}\text{C}$ respectively
C. $80circC$ $80^{circ}\text{C}$ and $70circC$ $70^{circ}\text{C}$ respectively
D. $60circC$ $60^{circ}\text{C}$ and $45circC$ $45^{circ}\text{C}$ respectively

Solution

Related Formula

H = Δ TReq, R = (l)/(KA)

Core Logic

Solution thermal network for Q38 - JEE Main 2026 Morning — Composite rods x and y joined end-to-end with temperature endpoints.

Let thermal resistance of rod y be $3R$ $3R$ and rod x be $R$ $R$ (since $Kₓ = 3Ky$ $K_x = 3K_y$).

Using thermal current equations:

H = (100 - 40)/((11R)/(2))

H = (100 - TB)/(R), H = (TE - 40)/(3R)

Solving yields $TB = 89circC$ $T_B = 89^{circ}\text{C}$ and $TE = 73circC$ $T_E = 73^{circ}\text{C}$.

Pattern Recognition

Sees: Composite thermal conduction network with series/parallel segments. Shortcut: Calculate equivalent thermal resistances and equate heat currents across junctions. Check: Matches option (1). ✓

Chapter Mix

Class 11 Physics: Thermal Properties of Matter

Q26 jee_main_2026_28_january_morning Calorimetry and Heat Transfer

10 kg of ice at

-10°C

is added to

100kg

of water to lower its temperature from

25°C

. Consider no heat exchange to surroundings. The decrement to the temperature of water is ____ °C. (specific heat of ice =

2100 ~J/Kg.°C

, specific heat of water =

4200 ~J/Kg.°C

, latent heat of fusion of ice =

3.36 \times 10⁵ ~J/Kg

)

A. $10$ $10$
B. $15$ $15$
C. $6.67$ $6.67$
D. $11.6$ $11.6$

Solution

Related Formula

$Q = mcΔ T$ $Q = mc\Delta T$ $Q = mL$ $Q = mL$

Core Logic

Heat gained by ice to reach final temperature $T$ $T$ must equal the heat lost by the water to reach temperature $T$ $T$.

Step 1: Energy Balance Equation

Let $T$ $T$ be the final temperature of the mixture. Heat gained by ice to reach $0°C$ $0^{\circ}\mathrm{C}$ + Heat to melt + Heat to reach $T$ $T$ = Heat lost by water.

10 \times 3.36 \times 10⁵ + 10 \times 2100 \times 10 + 10 \times 4200 \times (T - 0) = 100 \times 4200 \times (25 - T)

Step 2: Solving for Final Temperature

3.36 \times 10⁶ + 2.1 \times 10⁵ + 4.2 \times 10⁴ T = 1.05 \times 10⁷ - 4.2 \times 10⁵ T

3.57 \times 10⁶ + 4.2 \times 10⁴ T = 1.05 \times 10⁷ - 4.2 \times 10⁵ T

Solving for $T$ $T$:

4.62 \times 10⁵ T = 6.93 \times 10⁶

T = 15°C

Step 3: Calculating Decrement

The decrement in the temperature of water is:

Δ T = 25 - 15 = 10°C

Pattern Recognition

When mixing ice and water, track the 3 stages for ice (warm to 0, melt, warm to T). Equate total heat gained to heat lost by water. Always solve for T first, then check what the question actually asks (decrement).

Chapter Mix

Class 11 Physics: Thermal Properties of Matter

Q33 jee_main_2026_28_january_morning Calorimetry and Heat Transfer

Which of the following best represents the temperature versus heat supplied graph for water, in the range of

-20°C

120°C

A. Graph 1
B. Graph 2
C. Graph 3
D. Graph 4

Solution

Related Formula

$Q = mcΔ T$ $Q = mc\Delta T$ $Q = mL$ $Q = mL$

Core Logic

When heat is supplied steadily:

Ice warms from $-20°C$ $-20^{\circ}\mathrm{C}$ to $0°C$ $0^{\circ}\mathrm{C}$ (slope = $1/(m \cdot cice)$ $1/(m \cdot c_{\text{ice}})$).
Ice melts at $0°C$ $0^{\circ}\mathrm{C}$ (temperature constant, flat horizontal line).
Water warms from $0°C$ $0^{\circ}\mathrm{C}$ to $100°C$ $100^{\circ}\mathrm{C}$ (slope = $1/(m \cdot cwater)$ $1/(m \cdot c_{\text{water}})$).
Water boils at $100°C$ $100^{\circ}\mathrm{C}$ (temperature constant, longer flat horizontal line because $Lv > Lf$ $L_v > L_f$).
Steam warms from $100°C$ $100^{\circ}\mathrm{C}$ to $120°C$ $120^{\circ}\mathrm{C}$.

Step 1: Identifying Graph Features

The graph must have two flat plateaus corresponding to phase changes at $0°C$ $0^{\circ}\mathrm{C}$ and $100°C$ $100^{\circ}\mathrm{C}$. The heating curves (slanted lines) should show the temperature rising. Option (2) perfectly matches these distinct segments.

Pattern Recognition

Look for the horizontal phase-change steps precisely at 0 and 100 degrees Celsius for standard pressure $H₂O$ $\mathrm{H_2O}$.

Chapter Mix

Class 11 Physics: Thermal Properties of Matter

Solution & Explanation

Related Formula

Core Logic

Step 1: Solving for Gas Consumption

Pattern Recognition

Chapter Mix

Reference Study Guides

More Thermal Properties of Matter Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Calculate Final Length

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Energy Balance Equation

Step 2: Solving for Final Temperature

Step 3: Calculating Decrement

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Identifying Graph Features

Pattern Recognition

Chapter Mix

More Thermal Properties of Matter Questions — jee_main_2026_21_jan_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back