A calorimeter of water equivalent 20 {g} contains 180 {g} of water at 25^\circ{C} . 'm' grams of steam at 100^\circ{C} is mixed in it till the temperature of the mixure is 31^\circ{C}. The value of 'm' is close to (Latent heat of water =540 cal {g}^{-1}, specific heat of water =1 cal {g}^{-1}{ } ^\circ{C}^{-1} ) [3 Sep 2020 Shift 2]

JEE Mains 03-Sep-2020 Shift 2 Solved Paper

Section: Physics

Question : 19 of 75

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A calorimeter of water equivalent 20 g contains 180 g of water at 25°C. 'm' grams of steam at 100°C is mixed in it till the temperature of the mixure is 31°C. The value of 'm' is close to (Latent heat of water =540 cal g−1, specific heat of water =1 cal g−1‌ °C−1)

[3 Sep 2020 Shift 2]

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