A gas based geyser heats water flowing at the rate of 5.0$5.0$ litres per minute from 27^circtextC$27^{\circ}\text{C}$ to 87^circtextC$87^{\circ}\text{C}$.
The rate of consumption of the gas is ____ g/s.
(Take heat of combustion of gas = 5.0 times 10^4$5.0 \times 10^{4}$ J/g, specific heat capacity of water = 4200text J/kg.^circtextC$4200\text{ J/kg.}^{\circ}\text{C}$)
A.2.1
B.4.2
C.0.42
D.0.21
Solution & Explanation
### Related Formula
P = fracdmdt S Delta T$$P = \frac{dm}{dt} S \Delta T$$P_textheater = textRate of combustion times textHeat of combustion$$P_{\text{heater}} = \text{Rate of combustion} \times \text{Heat of combustion}$$
### Core Logic
Water flow rate = 5text l/min = frac560text kg/s = frac112text kg/s$5\text{ l/min} = \frac{5}{60}\text{ kg/s} = \frac{1}{12}\text{ kg/s}$.
The power required to heat the water is:
P_textheater = left(fracdmdtright)_textwater cdot S cdot Delta T$$P_{\text{heater}} = \left(\frac{dm}{dt}\right)_{\text{water}} \cdot S \cdot \Delta T$$P_textheater = frac112 times 4200 times (87 - 27) = frac112 times 4200 times 60text W$$P_{\text{heater}} = \frac{1}{12} \times 4200 \times (87 - 27) = \frac{1}{12} \times 4200 \times 60\text{ W}$$
### Step 1: Solving for Gas Consumption
Let the rate of consumption of gas be xtext g/s$x\text{ g/s}$.
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$$x \times 5.0 \times 10^{4} = \frac{1}{12} \times 4200 \times 60$$x times 5.0 times 10^4 = 4200 times 5$$x \times 5.0 \times 10^{4} = 4200 \times 5$$x = frac4200 times 55 times 10^4 = 4200 times 10^-4$$x = \frac{4200 \times 5}{5 \times 10^{4}} = 4200 \times 10^{-4}$$x = 0.42text g/s$$x = 0.42\text{ g/s}$$Calorimetry diagram for Q27 - JEE Main 2026 Morning
### Pattern Recognition
Equate the thermal power absorbed by fluid (dm/dt cdot c cdot Delta T$dm/dt \cdot c \cdot \Delta T$) with the power output of fuel (dm_textfuel/dt cdot L_textcombustion$dm_{\text{fuel}}/dt \cdot L_{\text{combustion}}$).
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
Keywords:#gas based geyser heats water flowing#JEE Main 2026 Morning Q27#Thermal Properties of Matter JEE Main 2026#Calorimetry JEE Main 2026
More Thermal Properties of Matter Previous-Year Questions
Q40jee_main_2026_21_jan_morningThermal 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/^circmathrmC$24 \times 10^{-6}/^{\circ}\mathrm{C}$ and 1.2 times 10^-5/^circmathrmC$1.2 \times 10^{-5}/^{\circ}\mathrm{C}$, 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
Delta l = l_0 alpha Delta T$$\Delta l = l_0 \alpha \Delta T$$l_textfinal = l_1 + l_2 + Delta l_1 + Delta l_2$$l_{\text{final}} = l_1 + l_2 + \Delta l_1 + \Delta l_2$$
### Core Logic
Let the original lengths be l_0A$l_{0A}$ (aluminium) and l_0S$l_{0S}$ (steel).
They have same length, so l_0A = l_0S = 60text cm$l_{0A} = l_{0S} = 60\text{ cm}$.
Change in temperature Delta T = 100^circtextC - 30^circtextC = 70^circtextC$\Delta T = 100^{\circ}\text{C} - 30^{\circ}\text{C} = 70^{\circ}\text{C}$.
Expansion of Aluminium: Delta l_A = l_0 alpha_A Delta T$\Delta l_A = l_0 \alpha_A \Delta T$
Expansion of Steel: Delta l_S = l_0 alpha_S Delta T$\Delta l_S = l_0 \alpha_S \Delta T$
### Step 1: Calculate Final Length
l_textfinal = l_0(1 + alpha_A Delta T) + l_0(1 + alpha_S Delta T)$$l_{\text{final}} = l_0(1 + \alpha_A \Delta T) + l_0(1 + \alpha_S \Delta T)$$l_textfinal = l_0 [2 + (alpha_A + alpha_S)Delta T]$$l_{\text{final}} = l_0 [2 + (\alpha_A + \alpha_S)\Delta T]$$l_textfinal = 60 left[ 2 + (24 times 10^-6 + 12 times 10^-6) times 70 right]$$l_{\text{final}} = 60 \left[ 2 + (24 \times 10^{-6} + 12 \times 10^{-6}) \times 70 \right]$$l_textfinal = 60 left[ 2 + (36 times 10^-6) times 70 right]$$l_{\text{final}} = 60 \left[ 2 + (36 \times 10^{-6}) \times 70 \right]$$l_textfinal = 60 [2 + 0.00252] = 120 + 60(0.00252) = 120 + 0.1512 = 120.15text cm$$l_{\text{final}} = 60 [2 + 0.00252] = 120 + 60(0.00252) = 120 + 0.1512 = 120.15\text{ cm}$$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.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
Q19jee_main_2025_02_april_eveningSpecific Heat Capacity and Thermal Conductivity
Match List-I with List-II.
List-I:
(A) Heat capacity of body
(B) Specific heat capacity of body
(C) Latent heat
(D) Thermal conductivity
List-II:
(I) mathrmJkg^-1$\mathrm{Jkg^{-1}}$
(II) mathrmJK^-1$\mathrm{JK}^{-1}$
(III) mathrmJkg^-1K^-1$\mathrm{Jkg^{-1}K^{-1}}$
(IV) mathrmJm^-1mathrmK^-1mathrms^-1$\mathrm{Jm}^{-1}\mathrm{K}^{-1}\mathrm{s}^{-1}$
Choose the correct answer from the options given below:
### Related Formula
Governing mathematical equations for thermal properties:
1. Heat capacity: C' = fracDelta QDelta T$C' = \frac{\Delta Q}{\Delta T}$
2. Specific heat capacity: s = fracDelta Qm Delta T$s = \frac{\Delta Q}{m \Delta T}$
3. Latent heat: L = fracDelta Qm$L = \frac{\Delta Q}{m}$
4. Thermal conductivity: K = fracDelta Q cdot LA cdot Delta T cdot t$K = \frac{\Delta Q \cdot L}{A \cdot \Delta T \cdot t}$
### Core Logic
Let's derive the SI units for each property:
- **(A) Heat capacity of body:**
[C'] = fracmathrmJmathrmK = mathrmJK^-1 quad implies text(II)$$[C'] = \frac{\mathrm{J}}{\mathrm{K}} = \mathrm{JK^{-1}} \quad \implies \text{(II)}$$
- **(B) Specific heat capacity of body:**
[s] = fracmathrmJmathrmkg cdot K = mathrmJkg^-1K^-1 quad implies text(III)$$[s] = \frac{\mathrm{J}}{\mathrm{kg \cdot K}} = \mathrm{Jkg^{-1}K^{-1}} \quad \implies \text{(III)}$$
- **(C) Latent heat:**
[L] = fracmathrmJmathrmkg = mathrmJkg^-1 quad implies text(I)$$[L] = \frac{\mathrm{J}}{\mathrm{kg}} = \mathrm{Jkg^{-1}} \quad \implies \text{(I)}$$
- **(D) Thermal conductivity:**
From heat transfer equation fracdQdt = K A fracdTdx$\frac{dQ}{dt} = K A \frac{dT}{dx}$:
K = fracdQ/dtA (dT/dx) implies [K] = fracmathrmJ/smathrmm^2 cdot (mathrmK/m) = mathrmJ cdot m^-1 cdot K^-1 cdot s^-1 quad implies text(IV)$$K = \frac{dQ/dt}{A (dT/dx)} \implies [K] = \frac{\mathrm{J/s}}{\mathrm{m}^2 \cdot (\mathrm{K/m})} = \mathrm{J \cdot m^{-1} \cdot K^{-1} \cdot s^{-1}} \quad \implies \text{(IV)}$$
### Step 1: Match values
Applying our mappings:
- (A) to$\to$ (II)
- (B) to$\to$ (III)
- (C) to$\to$ (I)
- (D) to$\to$ (IV)
This maps to Option (4).
### Pattern Recognition
Sees: Dimensional unit matching of thermal parameters.
Trap: Confusing Specific Heat Capacity (scaled per mass unit) with Heat Capacity (unscaled entire body parameter).
Shortcut: Specific heat is normalized by mass, meaning its unit must contain kg in the denominator. Heat capacity has no mass constraint, meaning A maps directly to II.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
Q24jee_main_2025_07_april_morningRadiation
A wire of length 10mathrmcm$10\mathrm{cm}$ and diameter 0.5mathrmmm$0.5\mathrm{mm}$ is used in a bulb. The temperature of the wire is 1727^circmathrmC$1727^{\circ}\mathrm{C}$ and power radiated by the wire is 94.2 W. Its emissivity is fracmathrmx8$\frac{\mathrm{x}}{8}$ where mathrmx =$\mathrm{x} =$
(Given sigma = 6.0 times 10^-8 mathrm~W mathrm~m^-2 mathrm~K^-4$\sigma = 6.0 \times 10^{-8} \mathrm{~W} \mathrm{~m}^{-2} \mathrm{~K}^{-4}$ , pi = 3.14$pi = 3.14$ and assume that the emissivity of wire material is same at all wavelength.)
Numerical Answer.Answer: 5 to 5
Solution
### Related Formula
Stefan-Boltzmann Law for radiated thermal power:
P = epsilon sigma A T^4$$P = \epsilon \sigma A T^4$$
Surface area A$A$ of a cylindrical wire of diameter d$d$ and length L$L$ is:
A = pi d L$A = \pi d L$
### Core Logic
Convert given parameters to standard SI units:
- L = 10 mathrm~cm = 0.1 mathrm~m$L = 10 \mathrm{~cm} = 0.1 \mathrm{~m}$
- d = 0.5 mathrm~mm = 0.5 times 10^-3 mathrm~m$d = 0.5 \mathrm{~mm} = 0.5 \times 10^{-3} \mathrm{~m}$
- T = 1727 + 273.15 = 2000 mathrm~K$T = 1727 + 273.15 = 2000 \mathrm{~K}$
- P = 94.2 mathrm~W$P = 94.2 \mathrm{~W}$
- sigma = 6.0 times 10^-8 mathrm~W cdot m^-2 cdot K^-4$\sigma = 6.0 \times 10^{-8} \mathrm{~W \cdot m^{-2} \cdot K^{-4}}$
### Step 1: Express Radiated Power and Emissivity
First, calculate the surface area A$A$:
A = 3.14 times (0.5 times 10^-3) times 0.1 = 1.57 times 10^-4 mathrm~m^2$$A = 3.14 \times (0.5 \times 10^{-3}) \times 0.1 = 1.57 \times 10^{-4} \mathrm{~m}^2$$
Substitute A$A$, sigma$\sigma$, and T$T$ into Stefan's formula:
94.2 = epsilon times (6.0 times 10^-8) times [3.14 times (0.5 times 10^-3) times (10 times 10^-2)] times (2000)^4$$94.2 = \epsilon \times (6.0 \times 10^{-8}) \times [3.14 \times (0.5 \times 10^{-3}) \times (10 \times 10^{-2})] \times (2000)^4$$
Calculate temperature term:
(2000)^4 = 1.6 times 10^12$$(2000)^4 = 1.6 \times 10^{12}$$94.2 = epsilon times (6 times 10^-8) times (1.57 times 10^-4) times (16 times 10^12)$$94.2 = \epsilon \times (6 \times 10^{-8}) \times (1.57 \times 10^{-4}) \times (16 \times 10^{12})$$94.2 = epsilon times 6 times 1.57 times 16 times 1 = epsilon times 150.72$$94.2 = \epsilon \times 6 \times 1.57 \times 16 \times 1 = \epsilon \times 150.72$$epsilon = frac94.2150.72 = frac58$$\epsilon = \frac{94.2}{150.72} = \frac{5}{8}$$
Thus, fracx8 = frac58 implies x = 5$\frac{x}{8} = \frac{5}{8} \implies x = 5$.
### Pattern Recognition
Sees: Bulb filament heat radiation equation.
Shortcut: Simplify the multiplication with factors of 10 first. T=2000 mathrm~K$T=2000 \mathrm{~K}$ has four zeros, so T^4$T^4$ adds 10^12$10^{12}$ which cancels the 10^-8$10^{-8}$ and 10^-4$10^{-4}$ from area and constant. The coefficient equation directly yields the ratio epsilon = 5/8$\epsilon = 5/8$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
Q19jee_main_2025_29_jan_eveningNewton's Law of Cooling
A cup of coffee cools from 90^circmathrmC$90^{\circ}\mathrm{C}$ to 80^circmathrmC$80^{\circ}\mathrm{C}$ in t$t$ minutes when the room temperature is 20^circmathrmC$20^{\circ}\mathrm{C}$ . The time taken by the similar cup of coffee to cool from 80^circmathrmC$80^{\circ}\mathrm{C}$ to 60^circmathrmC$60^{\circ}\mathrm{C}$ at the same room temperature is :
A.frac135 t$\frac{13}{5} t$
B.frac1013 t$\frac{10}{13} t$
C.frac1310 t$\frac{13}{10} t$
D.frac513 t$\frac{5}{13} t$
Solution
### Related Formula
fracT_i - T_fDelta t = Kleft(fracT_i + T_f2 - T_0right)$$\frac{T_i - T_f}{\Delta t} = K\left(\frac{T_i + T_f}{2} - T_0\right)$$
### Core Logic
**Case 1** (90^circmathrmC rightarrow 80^circmathrmC$90^{\circ}\mathrm{C} \rightarrow 80^{\circ}\mathrm{C}$ in time t$t$):
frac90 - 80t = Kleft(frac90 + 802 - 20
ight)$$\frac{90 - 80}{t} = K\left(\frac{90 + 80}{2} - 20
ight)$$frac10t = K(85 - 20) = 65K quad dots (i)$$\frac{10}{t} = K(85 - 20) = 65K \quad \dots (i)$$
**Case 2** (80^circmathrmC rightarrow 60^circmathrmC$80^{\circ}\mathrm{C} \rightarrow 60^{\circ}\mathrm{C}$ in time t'$t'$):
frac80 - 60t' = Kleft(frac80 + 602 - 20
ight)$$\frac{80 - 60}{t'} = K\left(\frac{80 + 60}{2} - 20
ight)$$frac20t' = K(70 - 20) = 50K quad dots (ii)$$\frac{20}{t'} = K(70 - 20) = 50K \quad \dots (ii)$$
Dividing equation (i)$(i)$ by equation (ii)$(ii)$:
frac10/t20/t' = frac65K50K$$\frac{10/t}{20/t'} = \frac{65K}{50K}$$fract'2t = frac1310 implies t' = frac135t$$\frac{t'}{2t} = \frac{13}{10} \implies t' = \frac{13}{5}t$$
### Pattern Recognition
Newton's law of cooling in average form uses the arithmetic mean of temperatures to approximate the driving temperature difference over the interval.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Physics: Thermal Properties of Matter
More Thermal Properties of Matter Questions — jee_main_2026_21_jan_morning
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