Water of volume 2 L in a closed… - JEE Main 2014 Physics PYQ

JEE Main 2014 — Physics Thermodynamics

2014mcqmedium

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 27 $^{o}$ C to 77 $^{o}$ C ? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible).

Official previous-year question

Held on 9 Apr 2014 · Verified 6 Jul 2026.

Options

A
8 min 20 s
B
7 min
C
14 min
D
6 min 2 s

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Solution

$\text{P} = 1 0 0 0 \text{ watt}$

$\therefore {\text{P}}_{\text{in}}=(1000-160)\text{J/s}=840\text{ J/s}$

Now, $\frac{ \text{d} \text{Q} }{ \text{dt} } = \text{ms} \frac{ \text{dT} }{ \text{dt} }$

$\therefore 840=\text{ms}\frac{\text{dT}}{\text{dt}}$

$\therefore \text{dt} = \frac{ \text{ms} }{ 8 4 0 } \text{dT}$

Integrating both sides we have;

$\text{t} = \frac{ \text{ms} }{ 8 4 0 } \int \text{dT}$

$= \frac{ ( 2 \times 1 0 0 0 \times 1 {0}^{ - 6 } \times 1 0 0 0 ) \times \text{4.2} \times 1 0 0 0 }{ 8 4 0 } \times 5 0$

$= 5 0 0 sec$

$= 8 min 2 0 sec$

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Solution

$\text{P} = 1 0 0 0 \text{ watt}$

$\therefore {\text{P}}_{\text{in}}=(1000-160)\text{J/s}=840\text{ J/s}$

Now, $\frac{ \text{d} \text{Q} }{ \text{dt} } = \text{ms} \frac{ \text{dT} }{ \text{dt} }$

$\therefore 840=\text{ms}\frac{\text{dT}}{\text{dt}}$

$\therefore \text{dt} = \frac{ \text{ms} }{ 8 4 0 } \text{dT}$

Integrating both sides we have;

$\text{t} = \frac{ \text{ms} }{ 8 4 0 } \int \text{dT}$

$= \frac{ ( 2 \times 1 0 0 0 \times 1 {0}^{ - 6 } \times 1 0 0 0 ) \times \text{4.2} \times 1 0 0 0 }{ 8 4 0 } \times 5 0$

$= 5 0 0 sec$

$= 8 min 2 0 sec$

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Options

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

Keep practicing Thermodynamics