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

Solution & Explanation

### Related Formula

P = fracdmdt S Delta T

P_textheater = textRate of combustion times textHeat of combustion

### Core Logic Water flow rate =

5text l/min = frac560text kg/s = frac112text kg/s

. The power required to heat the water is:

P_textheater = left(fracdmdtright)_textwater cdot S cdot Delta T

P_textheater = frac112 times 4200 times (87 - 27) = frac112 times 4200 times 60text W

### Step 1: Solving for Gas Consumption Let the rate of consumption of gas be

xtext g/s

. Heat generated by the gas per second must equal the power required to heat the water:

x times 5.0 times 10^4 = frac112 times 4200 times 60

x times 5.0 times 10^4 = 4200 times 5

x = frac4200 times 55 times 10^4 = 4200 times 10^-4

x = 0.42text g/s

Calorimetry diagram for Q27 - JEE Main 2026 Morning

### Pattern Recognition Equate the thermal power absorbed by fluid (

dm/dt cdot c cdot Delta T

) with the power output of fuel (

dm_textfuel/dt cdot L_textcombustion

). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter

More Thermal Properties of Matter Previous-Year Questions — Page 2

Q3 jee_main_2025_04_april_evening Thermal Expansion

Consider a rectangular sheet of solid material of length

ell=9

cm and width

d=4

cm. The coefficient of linear expansion is

alpha=3.1times10^-5text K^-1

at room temperature and one atmospheric pressure. The mass of sheet

m=0.1text kg

and the specific heat capacity

C_v=900text J kg^-1textK^-1

. If the amount of heat supplied to the material is

8.1times10^2

J then change in area of the rectangular sheet is :-

A. $2.0times10^-6text m^2$ $2.0\times10^{-6}\text{ m}^{2}$
B. $3.0times10^-7text m^2$ $3.0\times10^{-7}\text{ m}^{2}$
C. $6.0times10^-7text m^2$ $6.0\times10^{-7}\text{ m}^{2}$
D. $4.0times10^-7text m^2$ $4.0\times10^{-7}\text{ m}^{2}$

Solution

### Related Formula

Delta Q = m C_v Delta T

Delta A = A_0 beta Delta T = A_0 (2alpha) Delta T

### Core Logic First, calculate the temperature change using heat supplied:

Delta T = fracDelta Qm C_v

Substituting the values:

Delta T = frac8.1 times 10^20.1 times 900 = frac81090 = 9text K

### Step 1: Calculate Change in Area Initial area

A_0 = ell times d = 9text cm times 4text cm = 36text cm^2 = 36 times 10^-4text m^2

. Now, substitute into the area expansion formula:

Delta A = 36 times 10^-4 times 2 times (3.1 times 10^-5) times 9

Delta A = 36 times 18 times 3.1 times 10^-9 = 2008.8 times 10^-9text m^2 approx 2.0 times 10^-6text m^2

### Pattern Recognition Always remember that areal expansion coefficient

beta = 2alpha

. Convert geometry dimensions to standard units (

1text cm^2 = 10^-4text m^2

) cleanly before concluding arithmetic. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter

Q23 jee_main_2025_07_april_evening Thermal Conduction

Two cylindrical rods A and B made of different materials, are joined in a straight line. The ratio of lengths, radii and thermal conductivities of these rods are:

fracmathrmL_mathrmAmathrmL_mathrmB = frac12, fracmathrmr_mathrmAmathrmr_mathrmB = 2

and

fracmathrmK_mathrmAmathrmK_mathrmB = frac12

. The free ends of rods A and B are maintained at

400mathrm~K

200mathrm~K

, respectively. The temperature of rods interface is ________ K, when equilibrium is established. [cite: 187, 188, 189, 190, 191, 192, 193]

Numerical Answer. Answer: 360 to 360

Solution

### Related Formula

R_textth = fracLKA = fracLK(pi r^2)

[cite: 817]

fracdQdt = fracDelta TR_textth

[cite: 818] ### Core Logic At steady state equilibrium, the rate of heat flow through both sections in series must be identical: [cite: 193, 819]

frac400 - TR_1 = fracT - 200R_2 implies frac400 - TT - 200 = fracR_1R_2

[cite: 192, 820] Let's evaluate the resistance ratio

fracR_1R_2

using the dimensional parameters: [cite: 820]

fracR_1R_2 = left(fracL_AL_Bright) cdot left(fracr_Br_Aright)^2 cdot left(fracK_BK_Aright)

[cite: 820] Substitute the given values:

fracL_AL_B = frac12

fracr_Ar_B = 2 implies fracr_Br_A = frac12

, and

fracK_AK_B = frac12 implies fracK_BK_A = 2

[cite: 189, 190]:

fracR_1R_2 = frac12 times left(frac12right)^2 times 2 = frac14

[cite: 821] Now link this back into the temperature equation: [cite: 822]

frac400 - TT - 200 = frac14 implies 1600 - 4T = T - 200

[cite: 822, 823]

5T = 1800 implies T = 360\ textK

[cite: 824, 825] ### Pattern Recognition Thermal conduction processes behave exactly like electric current fields in series connections[cite: 818, 819]. Cross-sectional area scales squarely with radius parameters, which requires extra care during substitution. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter

Q10 jee_main_2025_24_jan_evening Temperature Scales

Which of the following figure represents the relation between Celsius and Fahrenheit temperatures?

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

Solution

### Related Formula

fracC5 = fracF - 329

### Core Logic Rearranging the conversion identity to express

C

as a function of

F

C = frac59F - frac1609

This is a straight line equation

y = mx + c

with: - Positive slope

m = frac59

- Negative

y

-intercept

c = -frac1609

(at

F=0

C approx -17.8^circmathrmC

) - Positive

x

-intercept at

C=0, F=32

This completely matches the layout shown in Graph (2).

Correct linear plot for Celsius vs Fahrenheit conversion Q10 — celsius fahrenheit graph, temperature conversion line, linear scaling

### Pattern Recognition

0^circmathrmC = 32^circmathrmF

. Therefore, the line must cross the positive side of the horizontal

F

axis when

C=0

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter

Q16 jee_main_2025_24_jan_evening Newton's Law of Cooling

The temperature of a body in air falls from

40^circ

C to

24^circ

C in 4 minutes. The temperature of the air is

16^circ

C. The temperature of the body in the next 4 minutes will be:

A. $frac143^circmathrmC$ $\frac{14}{3}^{\circ}\mathrm{C}$
B. $frac283^circmathrmC$ $\frac{28}{3}^{\circ}\mathrm{C}$
C. $frac563^circmathrmC$ $\frac{56}{3}^{\circ}\mathrm{C}$
D. $frac423^circmathrmC$ $\frac{42}{3}^{\circ}\mathrm{C}$

Solution

### Related Formula

fracT_1 - T_2t = Kleft[fracT_1 + T_22 - T_s ight]

### Core Logic For the first interval (

40^circmathrmC

24^circmathrmC

4\ mathrmmin

with

T_s = 16^circmathrmC

frac40 - 244 = Kleft[frac40 + 242 - 16 ight]

frac164 = K[32 - 16] implies 4 = 16K implies K = frac14

For the next 4 minutes, let the final temperature be

T

frac24 - T4 = frac14left[frac24 + T2 - 16 ight]

24 - T = frac24 + T2 - 16

40 - T = frac24 + T2

80 - 2T = 24 + T implies 3T = 56 implies T = frac563^circmathrmC

### Pattern Recognition Newton's law of cooling approximation works beautifully when temperature differences are small. Always compute

K

from the first step and substitute directly into the second. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter

Q34 jee_main_2024_30_january_evening Heating Curve of Water

A block of ice at

-10^circmathrmC

is slowly heated and converted to steam at

100^circmathrmC

. Which of the following curves represent the phenomenon qualitatively:

A. $textOption 1$ $\text{Option 1}$
B. $textOption 2$ $\text{Option 2}$
C. $textOption 3$ $\text{Option 3}$
D. $textOption 4$ $\text{Option 4}$

Solution

### Core Logic The heating curve traces temperature vs heat supplied. Stage 1: Ice at

-10^circmathrmC

is heated to

0^circmathrmC

(Temperature rises, solid phase). Stage 2: Ice melts at

0^circmathrmC

into water (Temperature remains constant until all ice melts). This is a horizontal plateau. Stage 3: Water is heated from

0^circmathrmC

100^circmathrmC

(Temperature rises, liquid phase). Stage 4: Water boils at

100^circmathrmC

into steam (Temperature remains constant). This is a second horizontal plateau. Option (4) correctly shows this sequential step-like graph. ### Pattern Recognition Heating curves always exhibit horizontal segments during phase changes (latent heat) where temperature remains constant. The slopes of the inclined regions depend on the specific heat capacities of the respective phases. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Thermal Properties of Matter

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