A gas based geyser heats water flowing at the rate of 5.0$5.0$ litres per minute from 27°C$27^{\circ}\text{C}$ to 87°C$87^{\circ}\text{C}$.
The rate of consumption of the gas is ____ g/s.
(Take heat of combustion of gas = 5.0 × 10⁴$5.0 \times 10^{4}$ J/g, specific heat capacity of water = 4200 J/kg.°C$4200\text{ J/kg.}^{\circ}\text{C}$)
A.2.1
B.4.2
C.0.42
D.0.21
Solution & Explanation
Related Formula
P = (dm)/(dt) S Δ T$$P = \frac{dm}{dt} S \Delta T$$Pheater = Rate of combustion × Heat of combustion$$P_{\text{heater}} = \text{Rate of combustion} \times \text{Heat of combustion}$$
Calorimetry diagram for Q27 - JEE Main 2026 Morning
Pattern Recognition
Equate the thermal power absorbed by fluid (dm/dt · c · Δ T$dm/dt \cdot c \cdot \Delta T$) with the power output of fuel (dmfuel/dt · Lcombustion$dm_{\text{fuel}}/dt \cdot L_{\text{combustion}}$).
Rearranging the conversion identity to express C$C$ as a function of F$F$:
C = (5)/(9)F - (160)/(9)$$C = \frac{5}{9}F - \frac{160}{9}$$
This is a straight line equation y = mx + c$y = mx + c$ with:
Positive slope m = (5)/(9)$m = \frac{5}{9}$
Negative y$y$-intercept c = -(160)/(9)$c = -\frac{160}{9}$ (at F=0$F=0$, C ≈ -17.8°C$C \approx -17.8^{\circ}\mathrm{C}$)
Positive x$x$-intercept at C=0, F=32$C=0, F=32$
This completely matches the layout shown in Graph (2). celsius fahrenheit graph, temperature conversion line, linear scaling
Pattern Recognition
0°C = 32°F$0^{\circ}\mathrm{C} = 32^{\circ}\mathrm{F}$. Therefore, the line must cross the positive side of the horizontal F$F$ axis when C=0$C=0$.
Chapter Mix
Class 11 Physics: Thermal Properties of Matter
Q16jee_main_2025_24_jan_eveningNewton's Law of Cooling
The temperature of a body in air falls from 40°$40^{\circ}$C to 24°$24^{\circ}$C in 4 minutes. The temperature of the air is 16°$16^{\circ}$C. The temperature of the body in the next 4 minutes will be:
For the next 4 minutes, let the final temperature be T$T$:
(24 - T)/(4) = (1)/(4)[(24 + T)/(2) - 16]$$\frac{24 - T}{4} = \frac{1}{4}\left[\frac{24 + T}{2} - 16\right]$$24 - T = (24 + T)/(2) - 16$$24 - T = \frac{24 + T}{2} - 16$$40 - T = (24 + T)/(2)$$40 - T = \frac{24 + T}{2}$$80 - 2T = 24 + T 3T = 56 T = (56)/(3)°C$$80 - 2T = 24 + T \implies 3T = 56 \implies T = \frac{56}{3}^{\circ}\mathrm{C}$$
Pattern Recognition
Newton's law of cooling approximation works beautifully when temperature differences are small. Always compute K$K$ from the first step and substitute directly into the second.
Chapter Mix
Class 11 Physics: Thermal Properties of Matter
Qjee_main_2024_30_january_eveningHeating Curve of Water
A block of ice at -10°C$-10^{\circ}\mathrm{C}$ is slowly heated and converted to steam at 100°C$100^{\circ}\mathrm{C}$. Which of the following curves represent the phenomenon qualitatively:
A.
B.
C.
D.
Solution
Core Logic
The heating curve traces temperature vs heat supplied.
Stage 1: Ice at -10°C$-10^{\circ}\mathrm{C}$ is heated to 0°C$0^{\circ}\mathrm{C}$ (Temperature rises, solid phase).
Stage 2: Ice melts at 0°C$0^{\circ}\mathrm{C}$ into water (Temperature remains constant until all ice melts). This is a horizontal plateau.
Stage 3: Water is heated from 0°C$0^{\circ}\mathrm{C}$ to 100°C$100^{\circ}\mathrm{C}$ (Temperature rises, liquid phase).
Stage 4: Water boils at 100°C$100^{\circ}\mathrm{C}$ 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.
Chapter Mix
Class 11 Physics: Thermal Properties of Matter
More Thermal Properties of Matter Questions — jee_main_2026_21_jan_morning
Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
