A calorimeter of water equivalent 20   g contains 180   g of water at 25 ° C . m ' '…

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 mixture 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)
  1. 2
  2. 4
  3. 3.2
  4. 2.6

Solution

20031-25=m×540+m169

1200=m609

m≈2

Asked in: JEE Main 2020 (03 Sep Shift 2)

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