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JEE Main 20203 SeptemberEvening Shift

Exam: JEE Main
Year: 2020
Date: 3 September
Shift: Evening Shift
Subject: Physics
Chapter: Thermodynamics

JEE Main 2020 · 3 September · Evening Shift · Physics · Thermodynamics

If minimum possible work is done by a refrigerator in converting...

If minimum possible work is done by a refrigerator in converting 100 grams of water at

0 ° C

to ice, how much heat (in calories) is released to the surroundings at temperature

27 ° C

(Latent heat of ice

= 80Cal / gram

) to the nearest integer?

Correct answer

0

Step-by-step solution

${Q}_{2} = mL = 8000$

$\dfrac{w}{{Q}_{2}} = \dfrac{{Q}_{1} - {Q}_{2}}{{Q}_{2}} = \dfrac{{T}_{1} - {T}_{2}}{{T}_{2}}$

$\dfrac{{Q}_{1} - 8000}{8000} = \dfrac{27}{273}$

${Q}_{1} = 8791$

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