JEE Main 2020 · PhysicsNumerical answerNumerical valueMediumMulti-step

JEE Main 3 September 2020, Shift 2, Physics Q23

Question 23 of 75 in this shift, Physics question 23 of 25.

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

0^\circ

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

27^\circ

C (Latent heat of ice

= 80

Cal/gram) to the nearest integer ?

Official answer

8791.00

NTA final key (Sep 2020).

Chapter: Thermodynamics
Topic: Heat Engines and Refrigerators
Idea tested: Carnot Refrigerator COP via Temperatures

Same topic in other shifts

All Heat Engines and Refrigerators questions

8 Apr 2023, Shift 2 · Q34Work done by a Carnot engine operating between temperatures $127^\circ$ C and $27^\circ$ C is 2 kJ. The amount of heat transferred to the…EasySingle correct
12 Apr 2023, Shift 1 · Q38An engine operating between the boiling and freezing points of water will have A. efficiency more than 27%. B. efficiency less than the…MediumOther
25 Jul 2022, Shift 2 · Q45Let $\eta_1$ is the efficiency of an engine at $\mathrm{T_1}$ = 447°C and $\mathrm{T_2}$ = 147°C while $\eta_2$ is the efficiency at…EasySingle correct
27 Jul 2022, Shift 2 · Q38In 1 $^{st}$ case, Carnot engine operates between temperatures 300 K and 100 K. In 2 $^{nd}$ case, as shown in the figure, a combination of…EasyDiagram basedHas a figure
28 Jul 2022, Shift 1 · Q37A Carnot engine has efficiency of 50%. If the temperature of sink is reduced by $40^\circ$ C, its efficiency increases by 30%. The…MediumSingle correct
30 Jul 2022, Shift 2 · Q7A Carnot engine is operating between the two reservoirs of temperature 227 $^\circ$ C and 127 $^\circ$ C respectively. If the engine…EasySingle correct

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.