JEE Main 2020 · PhysicsMultiple choiceSingle correctEasyCalculation

JEE Main 4 September 2020, Shift 1, Physics Q7

Question 7 of 75 in this shift, Physics question 7 of 25.

The specific heat of water = 4200 J kg

^{-1}

^{-1}

and the latent heat of ice =

3.4\times10^5

J kg

^{-1}

. 100 grams of ice at 0°C is placed in 200 g of water at 25°C. The amount of ice that will melt as the temperature of water reaches 0°C is close to (in grams) :

(1)69.3
(2)61.7Official answer
(3)64.6
(4)63.8

Official answer

Option 2

NTA final key (Sep 2020).

Chapter: Thermal Properties of Matter
Topic: Calorimetry
Idea tested: Latent Heat and Phase Change

Same idea in other shifts

Asked 4× in all

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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.