P=1000 watt
∴Pin=(1000−160)J/s=840 J/s
Now, dtdQ=msdtdT
∴840=msdtdT
∴dt=840msdT
Integrating both sides we have;
t=840ms∫dT
=840(2×1000×10−6×1000)×4.2×1000×50
=500sec
=8min20sec
JEE Main 2014 — Physics Thermodynamics
Water of volume 2 L in a closed container is heated with a coil of 1 kW. While water is heated, the container loses energy at a rate of 160 J/s. In how much time will the temperature of water rise from 27oC to 77oC ? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible).
Held on 9 Apr 2014 · Verified 6 Jul 2026.
8 min 20 s
7 min
14 min
6 min 2 s
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
10 mole of oxygen is heated at constant volume from $30^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$. The change in the internal energy of the gas is $\_\_\_\_$ cal. (The molecular specific heat of oxygen at constant pressure, $C_{P}=7 \mathrm{cal} / \mathrm{mol}.{ }^{\circ} \mathrm{C}$ and $\mathrm{R}=2 \mathrm{cal} / \mathrm{mol}.{ }^{\circ} \mathrm{C}$.)
An ideal gas at pressure $P$ and temperature $T$ is expanding such that $PT^3 = $ constant. The coefficient of volume expansion of the gas is _______.
In the following $p-V$ diagram the equation of state along the curved path is given by $(V-2)^{2}=4 a p$ where $a$ is a constant. The total work done in the closed path is 
The volume of an ideal gas increases 8 times and temperature becomes $(1 / 4)^{\text {th }}$ of initial temperature during a reversible change. If there is no exchange of heat in this process $(\Delta \mathrm{Q}=0)$ then identify the gas from the following options (Assuming the gases given in the options are ideal gases):
A certain gas is isothermally compressed to $\left(\dfrac{1}{3}\right)^{rd}$ of its initial volume ($V_o = 3$ litre) by applying required pressure. If the bulk modulus of the gas is $3 \times 10^5$ N/m$^2$, the magnitude of work done on the gas is _______ J.
Work through every JEE Main Thermodynamics PYQ, year by year.