(200)(31−25)=m×540+m(1)(69)
1200=m(609)
m≈2
JEE Main 2020 — Physics Thermodynamics
A calorimeter of water equivalent 20g contains 180g of water at 25∘C. m′′' grams of steam at 100∘C is mixed in it till the temperature of the mixture is 31∘C. The value of m′′ is close to (Latent heat of water=540calg−1, specific heat of water=1calg−1∘C−1)
Held on 3 Sept 2020 · Verified 6 Jul 2026.
2
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3.2
2.6
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